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Chemistry LibreTexts

5.1.1: Practice Problems- Writing and Balancing Chemical Equations

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PROBLEM \(\PageIndex{1}\)

Balance the following equations:

  • \(\ce{PCl5}(s)+\ce{H2O}(l)\rightarrow \ce{POCl3}(l)+\ce{HCl}(aq)\)
  • \(\ce{Cu}(s)+\ce{HNO3}(aq)\rightarrow \ce{Cu(NO3)2}(aq)+\ce{H2O}(l)+\ce{NO}(g)\)
  • \(\ce{H2}(g)+\ce{I2}(s)\rightarrow \ce{HI}(s)\)
  • \(\ce{Fe}(s)+\ce{O2}(g)\rightarrow \ce{Fe2O3}(s)\)
  • \(\ce{Na}(s)+\ce{H2O}(l)\rightarrow \ce{NaOH}(aq)+\ce{H2}(g)\)
  • \(\ce{(NH4)2Cr2O7}(s)\rightarrow \ce{Cr2O3}(s)+\ce{N2}(g)+\ce{H2O}(g)\)
  • \(\ce{P4}(s)+\ce{Cl2}(g)\rightarrow \ce{PCl3}(l)\)
  • \(\ce{PtCl4}(s)\rightarrow \ce{Pt}(s)+\ce{Cl2}(g)\)

\(\ce{PCl5}(s)+\ce{H2O}(l)\rightarrow \ce{POCl3}(l)+\ce{2HCl}(aq)\)

\(\ce{3Cu}(s)+\ce{8HNO3}(aq)\rightarrow \ce{3Cu(NO3)2}(aq)+\ce{4H2O}(l)+\ce{2NO}(g)\)

\(\ce{H2}(g)+\ce{I2}(s)\rightarrow \ce{2HI}(s)\)

\(\ce{4Fe}(s)+\ce{3O2}(g)\rightarrow \ce{2Fe2O3}(s)\)

\(\ce{2Na}(s)+\ce{2H2O}(l)\rightarrow \ce{2NaOH}(aq)+\ce{H2}(g)\)

\(\ce{(NH4)2Cr52O7}(s)\rightarrow \ce{Cr2O3}(s)+\ce{N2}(g)+\ce{4H2O}(g)\)

\(\ce{P4}(s)+\ce{6Cl2}(g)\rightarrow \ce{4PCl3}(l)\)

\(\ce{PtCl4}(s)\rightarrow \ce{Pt}(s)+\ce{2Cl2}(g)\)

PROBLEM \(\PageIndex{2}\)

  • \(\ce{Ag}(s)+\ce{H2S}(g)+\ce{O2}(g)\rightarrow \ce{Ag2S}(s)+\ce{H2O}(l)\)
  • \(\ce{P4}(s)+\ce{O2}(g)\rightarrow \ce{P4O10}(s)\)
  • \(\ce{Pb}(s)+\ce{H2O}(l)+\ce{O2}(g)\rightarrow \ce{Pb(OH)2}(s)\)
  • \(\ce{Fe}(s)+\ce{H2O}(l)\rightarrow \ce{Fe3O4}(s)+\ce{H2}(g)\)
  • \(\ce{Sc2O3}(s)+\ce{SO3}(l)\rightarrow \ce{Sc2(SO4)3}(s)\)
  • \(\ce{Ca3(PO4)2}(aq)+\ce{H3PO4}(aq)\rightarrow \ce{Ca(H2PO4)2}(aq)\)
  • \(\ce{Al}(s)+\ce{H2SO4}(aq)\rightarrow \ce{Al2(SO4)3}(s)+\ce{H2}(g)\)
  • \(\ce{TiCl4}(s)+\ce{H2O}(g)\rightarrow \ce{TiO2}(s)+\ce{HCl}(g)\)

\(\ce{4Ag}(s)+\ce{2H2S}(g)+\ce{O2}(g)\rightarrow \ce{2Ag2S}(s)+\ce{2H2O}(l)\)

\(\ce{P4}(s)+\ce{5O2}(g)\rightarrow \ce{P4O10}(s)\)

\(\ce{2Pb}(s)+\ce{2H2O}(l)+\ce{O2}(g)\rightarrow \ce{2Pb(OH)2}(s)\)

\(\ce{3Fe}(s)+\ce{4H2O}(l)\rightarrow \ce{Fe3O4}(s)+\ce{4H2}(g)\)

\(\ce{Sc2O3}(s)+\ce{3SO3}(l)\rightarrow \ce{Sc2(SO4)3}(s)\)

\(\ce{Ca3(PO4)2}(aq)+\ce{4H3PO4}(aq)\rightarrow \ce{3Ca(H2PO4)2}(aq)\)

\(\ce{2Al}(s)+\ce{3H2SO4}(aq)\rightarrow \ce{Al2(SO4)3}(s)+\ce{3H2}(g)\)

\(\ce{TiCl4}(s)+\ce{2H2O}(g)\rightarrow \ce{TiO2}(s)+\ce{4HCl}(g)\)

PROBLEM \(\PageIndex{3}\)

Write a balanced molecular equation describing each of the following chemical reactions.

  • Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
  • Gaseous butane, C 4 H 10 , reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.
  • Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
  • Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.

\(\ce{CaCO3}(s)\rightarrow \ce{CaO}(s)+\ce{CO2}(g)\)

\(\ce{2C4H10}(g)+\ce{13O2}(g)\rightarrow \ce{8CO2}(g)+\ce{10H2O}(g)\)

\(\ce{MgCl2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NaCl}(aq)\)

\(\ce{2H2O}(g)+\ce{2Na}(s)\rightarrow \ce{2NaOH}(s)+\ce{H2}(g)\)

PROBLEM \(\PageIndex{4}\)

Write a balanced equation describing each of the following chemical reactions.

  • Solid potassium chlorate, KClO 3 , decomposes to form solid potassium chloride and diatomic oxygen gas.
  • Solid aluminum metal reacts with solid diatomic iodine to form solid Al 2 I 6 .
  • When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.
  • Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.

\(\ce{2KClO3}(s)\rightarrow \ce{2KCl}(s)+\ce{3O2}(g)\)

\(\ce{2Al}(s)+\ce{3I2}(s)\rightarrow \ce{Al2I6}(s)\)

\(\ce{2NaCl}(s)+\ce{H2SO4}(aq)\rightarrow \ce{2HCl}(g)+\ce{Na2SO4}(aq)\)

\(\ce{H3PO4}(aq)+\ce{KOH}(aq)\rightarrow \ce{KH2PO4}(aq)+\ce{H2O}(l)\)

PROBLEM \(\PageIndex{5}\)

Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.

  • Write the formulas of barium nitrate and potassium chlorate.
  • The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.
  • The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.
  • Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains Fe 3+ ions.)

Ba(NO 3 ) 2 , KClO 3

\(\ce{2Ba(NO3)2}(s)\rightarrow \ce{2BaO}(s)+\ce{2N2}(g)+\ce{5O2}(g)\)

\(\ce{2Mg}(s)+\ce{O2}(g)\rightarrow \ce{2MgO}(s)\) ; \(\ce{4Al}(s)+\ce{3O2}(g)\rightarrow \ce{2Al2O3}(g)\); \(\ce{4Fe}(s)+\ce{3O2}(g)\rightarrow \ce{2Fe2O3}(s)\)

PROBLEM \(\PageIndex{6}\)

Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).

  • Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.
  • The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write the equation for this reaction.

\(\ce{4HF}(aq)+\ce{SiO2}(s)\rightarrow \ce{SiF4}(g)+\ce{2H2O}(l)\)

\(\ce{CaCl2}(aq)+\ce{2NaF}(aq)\rightarrow \ce{2NaCl}(aq)+\ce{CaF2}(s)\)

PROBLEM \(\PageIndex{7}\)

A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process.

  • The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.
  • The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.
  • Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.
  • The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.
  • Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.

\(\ce{CaO}(s)+\ce{H2O}(l)\rightarrow \ce{Ca(OH)2}(s)\)

\(\ce{Ca(OH)2}(s)+\ce{MgCl2}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{CaCl2}(aq)\)

\(\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)\rightarrow \ce{MgCl2}(aq)+\ce{2H2O}(l)\)

\(\ce{MgCl2}(s)\rightarrow \ce{Mg}(s)+\ce{Cl2}(g)\)

Contributors

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors.  Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).

  • Adelaide Clark, Oregon Institute of Technology

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Balancing Chemical Equations: Explanation, Review, and Examples

  • The Albert Team
  • Last Updated On: March 14, 2023

practice problems balancing chemical equations

Of all the skills to know about in chemistry, balancing chemical equations is perhaps the most important to master. So many parts of chemistry depend on this vital skill, including stoichiometry, reaction analysis, and lab work. This comprehensive guide will show you the steps to balance even the most challenging reactions and will walk you through a series of examples, from simple to complex.

The Key to Balancing Chemical Equations

The ultimate goal for balancing chemical equations is to make both sides of the reaction, the reactants and the products, equal in the number of atoms per element. This stems from the universal law of the conservation of mass, which states that matter can neither be created nor destroyed. So, if we start with ten atoms of oxygen before a reaction, we need to end up with ten atoms of oxygen after a reaction. This means that chemical reactions do not change the actual building blocks of matter; rather, they just change the arrangement of the blocks. An easy way to understand this is to picture a house made of blocks. We can break the house apart and build an airplane, but the color and shape of the actual blocks do not change.

But how do we go about balancing these equations? We know that the number of atoms of each element needs to be the same on both sides of the equation, so it is just a matter of finding the correct coefficients (numbers in front of each molecule) to make that happen. It is best to start with the atom that shows up the least number of times on one side, and balance that first. Then, move on to the atom that shows up the second least number of times, and so on. In the end, make sure to count the number of atoms of each element on each side again, just to be sure.

Example of Balancing a Chemical Equation

Let’s illustrate this with an example by balancing this chemical equation:

P 4 O 10 + H 2 O → H 3 PO 4

First, let’s look at the element that appears least often. Notice that oxygen occurs twice on the left-hand side, so that is not a good element to start out with. We could either start with phosphorus or hydrogen, so let’s start with phosphorus. There are four atoms of phosphorus on the left-hand side, but only one on the right-hand side. So, we can put the coefficient of 4 on the molecule that has phosphorous on the right-hand side to balance them out.

P 4 O 10 + H 2 O → 4 H 3 PO 4

Now we can check hydrogen. We still want to avoid balancing oxygen, because it occurs in more than one molecule on the left-hand side. It is easiest to start with molecules that only appear once on each side. So, there are two molecules of hydrogen on the left-hand side and twelve on the right-hand side (notice that there are three per molecule of H 3 PO 4 , and we have four molecules). So, to balance those out, we have to put a six in front of H 2 O on the left.

P 4 O 10 + 6 H 2 O → 4 H 3 PO 4

At this point, we can check the oxygens to see if they balance. On the left, we have ten atoms of oxygen from P 4 O 10 and six from H 2 O for a total of 16. On the right, we have 16 as well (four per molecule, with four molecules). So, oxygen is already balanced. This gives us the final balanced equation of

Balancing Chemical Equations Practice Problems

Try to balance these ten equations on your own, then check the answers below. They range in difficulty level, so don’t get discouraged if some of them seem too hard. Just remember to start with the element that shows up the least, and proceed from there. The best way to approach these problems is slowly and systematically. Looking at everything at once can easily get overwhelming. Good luck!

  • CO 2 + H 2 O → C 6 H 12 O 6 + O 2
  • SiCl 4 + H 2 O → H 4 SiO 4 + HCl
  • Al + HCl → AlCl 3 + H 2
  • Na 2 CO 3 + HCl → NaCl + H 2 O + CO 2
  • C 7 H 6 O 2 + O 2 → CO 2 + H 2 O
  • Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + Fe(OH) 3
  • Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + CaSiO 3
  • KClO 3 → KClO 4 + KCl
  • Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4
  • H 2 SO 4 + HI → H 2 S + I 2 + H 2 O

Complete Solutions:

1. co 2 + h 2 o → c 6 h 12 o 6 + o 2.

The first step to balancing chemical equations is to focus on elements that only appear once on each side of the equation. Here, both carbon and hydrogen fit this requirement. So, we will start with carbon. There is only one atom of carbon on the left-hand side, but six on the right-hand side. So, we add a coefficient of six to the carbon-containing molecule on the left.

6CO 2 + H 2 O → C 6 H 12 O 6 + O 2

Next, let’s look at hydrogen. There are two hydrogen atoms on the left and twelve on the right. So, we will add a coefficient of six on the hydrogen-containing molecule on the left.

6CO 2 + 6H 2 O → C 6 H 12 O 6 + O 2

Now, it is time to check the oxygen. There are a total of 18 oxygen molecules on the left (6×2 + 6×1). On the right, there are eight oxygen molecules. Now, we have two options to even out the right-hand side: We can either multiply C 6 H 12 O 6 or O 2 by a coefficient. However, if we change C 6 H 12 O 6 , the coefficients for everything else on the left-hand side will also have to change, because we will be changing the number of carbon and hydrogen atoms. To prevent this, it usually helps to only change the molecule containing the fewest elements; in this case, the O 2 . So, we can add a coefficient of six to the O 2 on the right. Our final answer will be:

6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6O 2

2. SiCl 4 + H 2 O → H 4 SiO 4 + HCl

The only element that occurs more than once on the same side of the equation here is hydrogen, so we can start with any other element. Let’s start by looking at silicon. Notice that there is only one atom of silicon on either side, so we do not need to add any coefficients yet. Next, let’s look at chlorine. There are four chlorine atoms on the left side and only one on the right. So, we will add a coefficient of four on the right.

SiCl 4 + H 2 O → H 4 SiO 4 + 4HCl

Next, let’s look at oxygen. Remember that we first want to analyze all the elements that only occur once on one side of the equation. There is only one oxygen atom on the left, but four on the right. So, we will add a coefficient of four on the left-hand side of the equation.

SiCl 4 + 4H 2 O → H 4 SiO 4 + 4HCl

We are almost done! Now, we just have to check the number of hydrogen atoms on each side. The left has eight and the right also has eight, so we are done. Our final answer is

As always, make sure to double-check that the number of atoms of each element balances on each side before continuing.

3. Al + HCl → AlCl 3 + H 2

This problem is a bit tricky, so be careful. Whenever a single atom is alone on either side of the equation, it is easiest to start with that element. So, we will start by counting the aluminum atoms on both sides. There is one on the left and one on the right, so we do not need to add any coefficients yet. Next, let’s look at hydrogen. There is also one on the left, but two on the right. So, we will add a coefficient of two on the left.

Al + 2HCl → AlCl 3 + H 2

Next, we will look at chlorine. There are now two on the left, but three on the right. Now, this is not as straightforward as just adding a coefficient to one side. We need the number of chlorine atoms to be equal on both sides, so we need to get two and three to be equal. We can accomplish this by finding the lowest common multiple. In this case, we can multiply two by three and three by two to get the lowest common multiple of six. So, we will multiply 2HCl by three and AlCl 3 by two:

Al + 6HCl → 2AlCl 3 + H 2

We have looked at all the elements, so it is easy to say that we are done. However, always make sure to double-check. In this case, because we added a coefficient to the aluminum-containing molecule on the right-hand side, aluminum is no longer balanced. There is one on the left but two on the right. So, we will add one more coefficient.

2Al + 6HCl → 2AlCl 3 + H 2

We are not quite done yet. Looking over the equation one final time, we see that hydrogen has also been unbalanced. There are six on the left but two on the right. So, with one final adjustment, we get our final answer:

2Al + 6HCl → 2AlCl 3 + 3H 2

4. Na 2 CO 3 + HCl → NaCl + H 2 O + CO 2

Hopefully, by this point, balancing equations is becoming easier and you are getting the hang of it. Looking at sodium, we see that it occurs twice on the left, but once on the right. So, we can add our first coefficient to the NaCl on the right.

Na 2 CO 3 + HCl → 2NaCl + H 2 O + CO 2

Next, let’s look at carbon. There is one on the left and one on the right, so there are no coefficients to add. Since oxygen occurs in more than one place on the left, we will save it for last. Instead, look at hydrogen. There is one on the left and two on the right, so we will add a coefficient to the left.

Na 2 CO 3 + 2HCl → 2NaCl + H 2 O + CO 2

Then, looking at chlorine, we see that it is already balanced with two on each side. Now we can go back to look at oxygen. There are three on the left and three on the right, so our final answer is

5. C 7 H 6 O 2 + O 2 → CO 2 + H 2 O

We can start balancing this equation by looking at either carbon or hydrogen. Looking at carbon, we see that there are seven atoms on the left and only one on the right. So, we can add a coefficient of seven on the right.

C 7 H 6 O 2 + O 2 → 7CO 2 + H 2 O

Then, for hydrogen, there are six atoms on the left and two on the right. So, we will add a coefficient of three on the right.

C 7 H 6 O 2 + O 2 → 7CO 2 + 3H 2 O

Now, for oxygen, things will get a little tricky. Oxygen occurs in every molecule in the equation, so we have to be very careful when balancing it. There are four atoms of oxygen on the left and 17 on the right. There is no obvious way to balance these numbers, so we must use a little trick: fractions. Now, when balancing chemical equations, we cannot include fractions as it is not proper form, but it sometimes helps to use them to solve the problem. Also, try to avoid over-manipulating organic molecules. You can easily identify organic molecules, otherwise known as CHO molecules, because they are made up of only carbon, hydrogen, and oxygen. We don’t like to work with these molecules, because they are rather complex. Also, larger molecules tend to be more stable than smaller molecules, and less likely to react in large quantities.

So, to balance out the four and seventeen, we can multiply the O 2 on the left by 7.5. That will give us

C 7 H 6 O 2 + 7.5O 2 → 7CO 2 + 3H 2 O

Remember, fractions (and decimals) are not allowed in formal balanced equations, so multiply everything by two to get integer values. Our final answer is now

2C 7 H 6 O 2 + 15O 2 → 14CO 2 + 6H 2 O

6. Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + Fe(OH) 3­-

We can start by balancing the iron on both sides. The left has two while the right only has one. So, we will add a coefficient of two to the right.

Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + 2Fe(OH) 3­-

Then, we can look at sulfur. There are three on the left, but only one on the right. So, we will add a coefficient of three to the right-hand side.

Fe 2 (SO 4 ) 3 + KOH → 3K 2 SO 4 + 2Fe(OH) 3­-

We are almost done. All that is left is to balance the potassium. There is one atom on the left and six on the right, so we can balance these by adding a coefficient of six. Our final answer, then, is

Fe 2 (SO 4 ) 3 + 6KOH → 3K 2 SO 4 + 2Fe(OH) 3­-

7. Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + CaSiO 3

Looking at calcium, we see that there are three on the left and one on the right, so we can add a coefficient of three on the right to balance them out.

Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 3CaSiO 3

Then, for phosphorus, we see that there are two on the left and four on the right. To balance these, add a coefficient of two on the left.

2Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 3CaSiO 3

Notice that by doing so, we changed the number of calcium atoms on the left. Every time you add a coefficient, double check to see if the step affects any elements you have already balanced. In this case, the number of calcium atoms on the left has increased to six while it is still three on the right, so we can change the coefficient on the right to reflect this change.

2Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 6CaSiO 3

Since oxygen occurs in every molecule in the equation, we will skip it for now. Focusing on silicon, we see that there is one on the left, but six on the right, so we can add a coefficient to the left.

2Ca 3 (PO 4 ) 2 + 6SiO 2 → P 4 O 10 + 6CaSiO 3

Now, we will check the number of oxygen atoms on each side. The left has 28 atoms and the right also has 28. So, after checking that all the other atoms are the same on both sides as well, we get a final answer of

8. KClO 3 → KClO 4 + KCl

This problem is particularly tricky because every atom, except oxygen, occurs in every molecule in the equation. So, since oxygen appears the least number of times, we will start there. There are three on the left and four on the right. To balance these, we find the lowest common multiple; in this case, 12. By adding a coefficient of four on the left and three on the right, we can balance the oxygens.

4KClO 3 → 3KClO 4 + KCl

Now, we can check potassium and chlorine. There are four potassium molecules on the left and four on the right, so they are balanced. Chlorine is also balanced, with four on each side, so we are finished, with a final answer of

9. Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4

We can start here by balancing the aluminum atoms on both sides. The left has two molecules while the right only has one, so we will add a coefficient of two on the right.

Al 2 (SO 4 ) 3 + Ca(OH) 2 → 2Al(OH) 3 + CaSO 4

Now, we can check sulfur. There are three on the left and only one on the right, so adding a coefficient of three will balance these.

Al 2 (SO 4 ) 3 + Ca(OH) 2 → 2Al(OH) 3 + 3CaSO 4

Moving right along to calcium, there is only one on the left but three on the right, so we should add a coefficient of three.

Al 2 (SO 4 ) 3 + 3Ca(OH) 2 → 2Al(OH) 3 + 3CaSO 4

Double-checking all the atoms, we see that all the elements are balanced, so our final equation is

10. H 2 SO 4 + HI → H 2 S + I 2 + H 2 O

Since hydrogen occurs more than once on the left, we will temporarily skip it and move to sulfur. There is one atom on the left and one on the right, so there is nothing to balance yet. Looking at oxygen, there are four on the left and one on the right, so we can add a coefficient of four to balance them.

H 2 SO 4 + HI → H 2 S + I 2 + 4H 2 O

There is only one iodine on the left and two on the right, so a simple coefficient change can balance those.

H 2 SO 4 + 2HI → H 2 S + I 2 + 4H 2 O

Now, we can look at the most challenging element: hydrogen. On the left, there are four and on the right, there are ten. So, we know we have to change the coefficient of either H 2 SO 4 or HI. We want to change something that will require the least amount of tweaking afterwards, so we will change the coefficient of HI. To get the left-hand side to have ten atoms of hydrogen, we need HI to have eight atoms of hydrogen, since H 2 SO 4 already has two. So, we will change the coefficient from 2 to 8.

H 2 SO 4 + 8HI → H 2 S + I 2 + 4H 2 O

However, this also changes the balance for iodine. There are now eight on the left, but only two on the right. To fix this, we will add a coefficient of 4 on the right. After checking that everything else balances out as well, we get a final answer of

H 2 SO 4 + 8HI → H 2 S + 4I 2 + 4H 2 O

As with most skills, practice makes perfect when balancing chemical equations. Keep working hard and try to do as many problems as you can to help you hone your balancing skills.

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  • 2 SnO₂ + 2 H₂ → 2 Sn + H₂O
  • SnO₂ + 2 H₂ → Sn + 2 H₂O
  • SnO₂ + H₂ → Sn + 2 H₂O
  • 2 SnO₂ + H₂ → 2Sn + H₂O
  • 3 KOH + 2 H₃PO₄ → 3 K₃PO₄ + 3 H₂O
  • KOH + H₃PO₄ → K₃PO₄ + 6 H₂O
  • 3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O
  • 2 KOH + H₃PO₄ → K₃PO₄ + H₂O
  • 2 KNO₃ + H₂CO₃ → K₂CO₃ + 2 HNO₃
  • 2 KNO₃ + 2 H₂CO₃ → K₂CO₃ + 2 HNO₃
  • KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃
  • 2 KNO₃ + 2 H₂CO₃ → K₂CO₃ + 3 HNO₃
  • AgI + Na₂S → Ag₂S + 2 NaI
  • 2 AgI + 2 Na₂S → 2 Ag₂S + 2 NaI
  • 2 AgI + 2 Na₂S → Ag₂S + 2 NaI
  • 2 AgI + Na₂S → Ag₂S + 2 NaI
  • 2 Ba₃N₂ + 3 H₂O → 2 Ba(OH)₂ + 2 NH₃
  • Ba₃N₂ + 6 H₂O → 3 Ba(OH)₂ + 2 NH₃
  • Ba₃N₂ + 3 H₂O → 3 Ba(OH)₂ + 2 NH₃
  • Ba₃N₂ + H₂O → Ba(OH)₂ + NH₃
  • 2 CaCl₂ + 2 Na₃PO₄ → 2 Ca₃(PO₄)₂ + NaCl
  • CaCl₂ + Na₃PO₄ → Ca₃(PO₄)₂ + NaCl
  • 3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 6 NaCl
  • 3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 3 NaCl
  • 2 FeS + O₂ → 2 Fe₂O₃ + 2 SO₂
  • FeS + O₂ → Fe₂O₃ + SO₂
  • 3 FeS + 2 O₂ → 3 Fe₂O₃ + 2 SO₂
  • 4 FeS + 7 O₂ → 2 Fe₂O₃ + 4 SO₂
  • 2 C₂H₆O + O₂ → CO₂ + 3 H₂O
  • C₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O
  • C₂H₆O + 2 O₂ → 3 CO₂ + 3 H₂O
  • 2 C₂H₆O + O₂ → 2 CO₂ + H₂O
  • TiCl₄ + 2 H₂O → TiO₂ + 2 HCl
  • TiCl₄ + 2 H₂O → TiO₂ + 4 HCl
  • 2 TiCl₄ + H₂O → 2 TiO₂ + HCl
  • TiCl₄ + 4 H₂O → TiO₂ + 4 HCl
  • Na₃PO₄ + HCl → NaCl + H₃PO₄
  • Na₃PO₄ +3 HCl → 3 NaCl + H₃PO₄
  • 3 Na₃PO₄ + HCl → 3 NaCl + H₃PO₄
  • Na₃PO₄ + 3 HCl → NaCl + H₃PO₄

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  • Balancing Equations Test Questions
  • How to Balance Chemical Equations
  • Balancing Chemical Equations
  • Examples of 10 Balanced Chemical Equations
  • Balanced Equation Definition and Examples
  • Mole Relations in Balanced Equations
  • 5 Steps for Balancing Chemical Equations
  • Equation for the Decomposition of Sodium Bicarbonate (Baking Soda)
  • Example Problem of Mass Relations in Balanced Equations
  • How to Balance Equations - Printable Worksheets
  • How to Balance Net Ionic Equations
  • Balance Redox Reaction Example Problem
  • Balance Redox Reaction in Basic Solution Example Problem
  • How to Balance Redox Reactions
  • Equilibrium Constants Practice Test
  • 4.1 Writing and Balancing Chemical Equations
  • Introduction
  • 1.1 Chemistry in Context
  • 1.2 Phases and Classification of Matter
  • 1.3 Physical and Chemical Properties
  • 1.4 Measurements
  • 1.5 Measurement Uncertainty, Accuracy, and Precision
  • 1.6 Mathematical Treatment of Measurement Results
  • Key Equations
  • 2.1 Early Ideas in Atomic Theory
  • 2.2 Evolution of Atomic Theory
  • 2.3 Atomic Structure and Symbolism
  • 2.4 Chemical Formulas
  • 2.5 The Periodic Table
  • 2.6 Ionic and Molecular Compounds
  • 2.7 Chemical Nomenclature
  • 3.1 Formula Mass and the Mole Concept
  • 3.2 Determining Empirical and Molecular Formulas
  • 3.3 Molarity
  • 3.4 Other Units for Solution Concentrations
  • 4.2 Classifying Chemical Reactions
  • 4.3 Reaction Stoichiometry
  • 4.4 Reaction Yields
  • 4.5 Quantitative Chemical Analysis
  • 5.1 Energy Basics
  • 5.2 Calorimetry
  • 5.3 Enthalpy
  • 6.1 Electromagnetic Energy
  • 6.2 The Bohr Model
  • 6.3 Development of Quantum Theory
  • 6.4 Electronic Structure of Atoms (Electron Configurations)
  • 6.5 Periodic Variations in Element Properties
  • 7.1 Ionic Bonding
  • 7.2 Covalent Bonding
  • 7.3 Lewis Symbols and Structures
  • 7.4 Formal Charges and Resonance
  • 7.5 Strengths of Ionic and Covalent Bonds
  • 7.6 Molecular Structure and Polarity
  • 8.1 Valence Bond Theory
  • 8.2 Hybrid Atomic Orbitals
  • 8.3 Multiple Bonds
  • 8.4 Molecular Orbital Theory
  • 9.1 Gas Pressure
  • 9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
  • 9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
  • 9.4 Effusion and Diffusion of Gases
  • 9.5 The Kinetic-Molecular Theory
  • 9.6 Non-Ideal Gas Behavior
  • 10.1 Intermolecular Forces
  • 10.2 Properties of Liquids
  • 10.3 Phase Transitions
  • 10.4 Phase Diagrams
  • 10.5 The Solid State of Matter
  • 10.6 Lattice Structures in Crystalline Solids
  • 11.1 The Dissolution Process
  • 11.2 Electrolytes
  • 11.3 Solubility
  • 11.4 Colligative Properties
  • 11.5 Colloids
  • 12.1 Chemical Reaction Rates
  • 12.2 Factors Affecting Reaction Rates
  • 12.3 Rate Laws
  • 12.4 Integrated Rate Laws
  • 12.5 Collision Theory
  • 12.6 Reaction Mechanisms
  • 12.7 Catalysis
  • 13.1 Chemical Equilibria
  • 13.2 Equilibrium Constants
  • 13.3 Shifting Equilibria: Le Châtelier’s Principle
  • 13.4 Equilibrium Calculations
  • 14.1 Brønsted-Lowry Acids and Bases
  • 14.2 pH and pOH
  • 14.3 Relative Strengths of Acids and Bases
  • 14.4 Hydrolysis of Salts
  • 14.5 Polyprotic Acids
  • 14.6 Buffers
  • 14.7 Acid-Base Titrations
  • 15.1 Precipitation and Dissolution
  • 15.2 Lewis Acids and Bases
  • 15.3 Coupled Equilibria
  • 16.1 Spontaneity
  • 16.2 Entropy
  • 16.3 The Second and Third Laws of Thermodynamics
  • 16.4 Free Energy
  • 17.1 Review of Redox Chemistry
  • 17.2 Galvanic Cells
  • 17.3 Electrode and Cell Potentials
  • 17.4 Potential, Free Energy, and Equilibrium
  • 17.5 Batteries and Fuel Cells
  • 17.6 Corrosion
  • 17.7 Electrolysis
  • 18.1 Periodicity
  • 18.2 Occurrence and Preparation of the Representative Metals
  • 18.3 Structure and General Properties of the Metalloids
  • 18.4 Structure and General Properties of the Nonmetals
  • 18.5 Occurrence, Preparation, and Compounds of Hydrogen
  • 18.6 Occurrence, Preparation, and Properties of Carbonates
  • 18.7 Occurrence, Preparation, and Properties of Nitrogen
  • 18.8 Occurrence, Preparation, and Properties of Phosphorus
  • 18.9 Occurrence, Preparation, and Compounds of Oxygen
  • 18.10 Occurrence, Preparation, and Properties of Sulfur
  • 18.11 Occurrence, Preparation, and Properties of Halogens
  • 18.12 Occurrence, Preparation, and Properties of the Noble Gases
  • 19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
  • 19.2 Coordination Chemistry of Transition Metals
  • 19.3 Spectroscopic and Magnetic Properties of Coordination Compounds
  • 20.1 Hydrocarbons
  • 20.2 Alcohols and Ethers
  • 20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
  • 20.4 Amines and Amides
  • 21.1 Nuclear Structure and Stability
  • 21.2 Nuclear Equations
  • 21.3 Radioactive Decay
  • 21.4 Transmutation and Nuclear Energy
  • 21.5 Uses of Radioisotopes
  • 21.6 Biological Effects of Radiation
  • A | The Periodic Table
  • B | Essential Mathematics
  • C | Units and Conversion Factors
  • D | Fundamental Physical Constants
  • E | Water Properties
  • F | Composition of Commercial Acids and Bases
  • G | Standard Thermodynamic Properties for Selected Substances
  • H | Ionization Constants of Weak Acids
  • I | Ionization Constants of Weak Bases
  • J | Solubility Products
  • K | Formation Constants for Complex Ions
  • L | Standard Electrode (Half-Cell) Potentials
  • M | Half-Lives for Several Radioactive Isotopes

Learning Objectives

By the end of this section, you will be able to:

  • Derive chemical equations from narrative descriptions of chemical reactions.
  • Write and balance chemical equations in molecular, total ionic, and net ionic formats.

An earlier chapter of this text introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation . Consider as an example the reaction between one methane molecule (CH 4 ) and two diatomic oxygen molecules (O 2 ) to produce one carbon dioxide molecule (CO 2 ) and two water molecules (H 2 O). The chemical equation representing this process is provided in the upper half of Figure 4.2 , with space-filling molecular models shown in the lower half of the figure.

This example illustrates the fundamental aspects of any chemical equation:

  • The substances undergoing reaction are called reactants , and their formulas are placed on the left side of the equation.
  • The substances generated by the reaction are called products , and their formulas are placed on the right side of the equation.
  • Plus signs (+) separate individual reactant and product formulas, and an arrow (⟶) (⟶) separates the reactant and product (left and right) sides of the equation.
  • The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.

It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the relative numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on ( Figure 4.3 ). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:

  • One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.
  • One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.
  • One mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide molecules and 2 moles of water molecules.

Balancing Equations

The chemical equation described in section 4.1 is balanced , meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO 2 and H 2 O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:

A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H 2 O to H 2 O 2 would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H 2 O to 2.

The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H 2 product to 2.

These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:

Example 4.1

Balancing chemical equations.

Next, count the number of each type of atom present in the unbalanced equation.

Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O 2 and N 2 O 5 to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).

The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N 2 to 2.

The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.

Check Your Learning

It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation’s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C 2 H 6 ) with oxygen to yield H 2 O and CO 2 , represented by the unbalanced equation:

Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:

This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O 2 reactant to yield an odd number, so a fractional coefficient, 7 2 , 7 2 , is used instead to yield a provisional balanced equation:

A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:

Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients . Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,

the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:

Link to Learning

Use this interactive tutorial for additional practice balancing equations.

Additional Information in Chemical Equations

The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include s for solids, l for liquids, g for gases, and aq for substances dissolved in water ( aqueous solutions , as introduced in the preceding chapter). These notations are illustrated in the example equation here:

This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).

Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation’s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (Δ) over the arrow.

Other examples of these special conditions will be encountered in more depth in later chapters.

Equations for Ionic Reactions

Given the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of CaCl 2 and AgNO 3 are mixed, a reaction takes place producing aqueous Ca(NO 3 ) 2 and solid AgCl:

This balanced equation, derived in the usual fashion, is called a molecular equation because it doesn’t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:

Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, s .

Explicitly representing all dissolved ions results in a complete ionic equation . In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:

Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, Ca 2+ ( aq ) and NO 3 − ( a q ) . NO 3 − ( a q ) . These spectator ions —ions whose presence is required to maintain charge neutrality—are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation :

Following the convention of using the smallest possible integers as coefficients, this equation is then written:

This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of Cl − and Ag + .

Example 4.2

Ionic and molecular equations.

Balance is achieved easily in this case by changing the coefficient for NaOH to 2, resulting in the molecular equation for this reaction:

The two dissolved ionic compounds, NaOH and Na 2 CO 3 , can be represented as dissociated ions to yield the complete ionic equation:

Finally, identify the spectator ion(s), in this case Na + ( aq ), and remove it from each side of the equation to generate the net ionic equation:

Write balanced molecular, complete ionic, and net ionic equations for this process.

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2FeCl 3 + MgO ---> Fe 2 O 3 + 3MgCl 2 The Fe also gets balanced in this step.
2FeCl 3 + 3MgO ---> Fe 2 O 3 + 3MgCl 2 The other element (Mg or O, depending on which one you picked) also gets balanced in this step.
3Li + H 3 PO 4 ---> H 2 + Li 3 PO 4
3Li + 2H 3 PO 4 ---> 3H 2 + Li 3 PO 4 Remember 2 x 3 = 6 and 3 x 2 = 6. It shows up a lot in balancing problems (if you haven't already figured that out!).
3Li + 2H 3 PO 4 ---> 3H 2 + 2Li 3 PO 4
6Li + 2H 3 PO 4 ---> 3H 2 + 2Li 3 PO 4
ZnS + (3/2)O 2 ---> ZnO + SO 2
2ZnS + 3O 2 ---> 2ZnO + 2SO 2
FeS 2 + (5/2)Cl 2 ---> FeCl 3 + S 2 Cl 2
2FeS 2 + 5Cl 2 ---> 2FeCl 3 + 2S 2 Cl 2
Fe + 3HC 2 H 3 O 2 ---> Fe(C 2 H 3 O 2 ) 3 + H 2
Fe + 3HC 2 H 3 O 2 ---> Fe(C 2 H 3 O 2 ) 3 + (3/2)H 2
2Fe + 6HC 2 H 3 O 2 ---> 2Fe(C 2 H 3 O 2 ) 3 + 3H 2
H 2 (g) + V 2 O 5 (s) ---> V 2 O 3 (s) + 2H 2 O(ℓ)
2H 2 (g) + V 2 O 5 (s) ---> V 2 O 3 (s) + 2H 2 O(ℓ)
4HCl(aq) + MnO 2 (s) ---> MnCl 2 (aq) + Cl 2 (g) + H 2 O(ℓ)
4HCl(aq) + MnO 2 (s) ---> MnCl 2 (aq) + Cl 2 (g) + 2H 2 O(ℓ)
Fe 2 O 3 (s) + C(s) ---> 2Fe(s) + CO 2 (g)
Fe 2 O 3 (s) + C(s) ---> 2Fe(s) + 3 ⁄ 2 CO 2 (g)
Fe 2 O 3 (s) + 3 ⁄ 2 C(s) ---> 2Fe(s) + 3 ⁄ 2 CO 2 (g)
2Fe 2 O 3 (s) + 3C(s) ---> 4Fe(s) + 3CO 2 (g)
2C 5 H 11 NH 2 + O 2 ---> CO 2 + 13H 2 O + NO 2
2C 5 H 11 NH 2 + O 2 ---> 10CO 2 + 13H 2 O + 2NO 2
2C 5 H 11 NH 2 + 37 ⁄ 2 O 2 ---> 10CO 2 + 13H 2 O + 2NO 2
4, 37 ---> 20, 26, 4
CO 2 + 3 ⁄ 8 S 8 ---> CS 2 + SO 2 Most of the time the fraction used to balance is something with a 2 in the denominator: 1 ⁄ 2 or 5 ⁄ 2 or 13 ⁄ 2 , for example. Not too often does one see 3 ⁄ 8 . Pretty tricky!
8CO 2 + 3S 8 ---> 8CS 2 + 8SO 2
P 4 + 3O 2 ---> 2P 2 O 3 This depends on seeing that the oxygen on the left comes in twos and the oxygen on the right comes in threes. So, you use a three and a two to arrive at six oxygens on each side. Least common multiple, baby!!
P 4 + O 2 ---> 2P 2 O 3 Then, the oxygen gets balanced: P 4 + 3O 2 ---> 2P 2 O 3
1 ⁄ 2 P 4 + O 2 ---> P 2 O 3 Let us balance the oxygen with a fraction as well: 1 ⁄ 2 P 4 + 3 ⁄ 2 O 2 ---> P 2 O 3 Finally, multiply through by two: P 4 + 3O 2 ---> 2P 2 O 3
  • Chemistry Concept Questions and Answers

Balanced Chemical Equations Questions

Chemical equations are symbolic representations of chemical reactions that express the reactants and products in terms of their respective chemical formulae. They also use symbols to represent factors such as reaction direction and the physical states of the reacting entities.

Balanced Chemical Equations Chemistry Questions with Solutions

Q1. A balanced chemical equation is in accordance with-

  • Multiple proportion
  • Reciprocal proportion
  • Conservation of mass
  • Definite proportions

Correct Answer: (c) Law of Conservation of Mass

Q2. The correct balanced equation for the reaction __C 2 H 6 O + __O 2 → __CO 2 + __H 2 O is-

  • 2C 2 H 6 O + O 2 → CO 2 + H 2 O
  • C 2 H 6 O + 3O 2 → 2CO 2 + 3H 2 O
  • C 2 H 6 O + 2O 2 → 3CO 2 + 3H 2 O
  • 2C 2 H 6 O + O 2 → 2CO 2 + H 2 O

Correct Answer: (b) C 2 H 6 O + 3O 2 → 2CO 2 + 3H 2 O

Q3. The correct balanced equation for the reaction __KNO 3 + __H 2 CO 3 → __K 2 CO 3 + __HNO 3 is-

  • 2KNO 3 + H 2 CO 3 → K 2 CO 3 + 2HNO 3
  • 2KNO 3 + 2H 2 CO 3 → K 2 CO 3 + 2HNO 3
  • KNO 3 + H 2 CO 3 → K 2 CO 3 + 2HNO 3
  • 2KNO 3 + 2H 2 CO 3 → K 2 CO 3 + 3HNO 3

Correct Answer: (a) 2KNO 3 + H 2 CO 3 → K 2 CO 3 + 2HNO 3

Q4. The correct balanced equation for the reaction __CaCl 2 + __Na 3 PO 4 → __ Ca 3 (PO 4 ) 2 + __ NaCl is-

  • 2CaCl 2 + 2Na 3 PO 4 → 2Ca 3 (PO 4 ) 2 + NaCl
  • CaCl 2 + Na 3 PO 4 → Ca 3 (PO 4 ) 2 + NaCl
  • 3CaCl 2 + 2Na 3 PO 4 → Ca 3 (PO 4 ) 2 + 6NaCl
  • 3CaCl 2 + 2Na 3 PO 4 → Ca 3 (PO 4 ) 2 + 3NaCl

Correct Answer- (c) 3CaCl 2 + 2Na 3 PO 4 → Ca3(PO 4 ) 2 + 6NaCl

Q5. The correct balanced equation for the reaction __TiCl 4 + __H 2 O → __TiO 2 + __HCl is-

  • TiCl 4 + 2H 2 O → TiO 2 +2HCl
  • TiCl 4 + 2H 2 O → TiO 2 + 4HCl
  • 2TiCl 4 + H 2 O → 2TiO 2 + HCl
  • TiCl 4 + 4H 2 O → TiO 2 + 4HCl

Correct Answer- (b) TiCl 4 + 2H 2 O → TiO 2 + 4HCl

Q6. Write a balanced equation for the reaction of molecular nitrogen (N 2 ) and oxygen (O 2 ) to form dinitrogen pentoxide.

Answer. The equation for the reaction is-

N 2 + O 2 → N 2 O 5 (unbalanced equation)

The balanced chemical equation is-

2N 2 + 5O 2 → 2N 2 O 5

Q7. On what basis is a chemical equation balanced?

Answer. A chemical equation is balanced using the law of conservation of mass, which states that “matter cannot be created nor destroyed.”

Q8. What is the balanced equation for the reaction of photosynthesis?

Answer. The balanced chemical equation for the reaction of photosynthesis is-

6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6O 2 .

Q9. We must solve a skeletal chemical equation.” Give a reason to justify the statement.

Answer. Skeletal chemical equations are unbalanced. Due to the law of conservation of mass, we must balance the chemical equation. It states that ‘matter cannot be created or destroyed.’ As a result, each chemical reaction must have a balanced chemical equation.

Q10. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?

Answer. The chemical equation must be balanced in order to obey the law of conservation of mass. A chemical equation is said to be balanced when the number of different atoms of elements in the reactants side equals the number of atoms in the products side. Balancing chemical equations is a trial-and-error process.

Q11. What is meant by the skeletal type chemical equation? What does it represent? Using the equation for electrolytic decomposition of water, differentiate between a skeletal chemical equation and a balanced chemical equation.

Answer. Skeletal equations are those in which formulas are used to indicate the chemicals involved in a chemical reaction.

The law of conservation of mass does not apply to skeletal equations.

The chemical formulas are represented by balanced chemical equations, which follow the law of conservation of mass, which states that the atoms on the reactant and product sides are the same.

H 2 O → H 2 + O 2 : Skeletal equation

2H 2 O → 2H 2 + O 2 : Balanced chemical equation

Q12. Write the balanced chemical equation for the following reaction:

  • Phosphorus burns in presence of chlorine to form phosphorus pentachloride.
  • Burning of natural gas.
  • The process of respiration.
  • P 4 + 10Cl 2 → 4PCl 5
  • CH 4 +2O 2 → CO 2 +2H 2 O + heat energy
  • C 6 H 12 O 6 + 6O 2 + 6H 2 O → 6CO 2 + 12H 2 O + energy

Q13. What Is the Distinction Between a Balanced Equation and a Skeleton Equation?

Answer. The primary distinction between a balanced equation and a skeleton equation is that the balanced equation provides the actual number of molecules of each reactant and product involved in the chemical reaction, whereas a skeleton equation only provides the reactants. Furthermore, a balanced equation may or may not contain stoichiometric coefficients, whereas a skeleton equation does not.

Q14. Balance the equations

  • HNO +Ca(OH) 2 → Ca(NO 3 ) 2 + H 2 O
  • NaCl + AgNO 3 → AgCl + NaNO 3
  • BaCl 2 +H 2 SO 4 → BaSO 4 +HCl

Answer. The balanced chemical equation for the reactions are as follows-

  • 2HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + 2H 2 O
  • BaCl 2 +H 2 SO 4 → BaSO 4 + 2HCl

Q15. Write a balanced molecular equation describing each of the following chemical reactions.

  • Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
  • Gaseous butane, C 4 H 10 , reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapour.
  • Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
  • Water vapour reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.
  • CaCO 3 → CaO + CO 2

On heating, 1 mol of solid calcium carbonate yields 1 mol of calcium oxide and 1 mol of carbon dioxide gas.

  • 2C 4 H 10 +13O 2 → 8CO 2 + 10H 2 O When 2 moles of butane gas react with 13 moles of diatomic oxygen gas, 8 moles of carbon dioxide gas and 10 moles of water vapours are produced.
  • MgCl 2 + 2NaOH → 2NaCl + Mg(OH) 2

1 mol magnesium Cordelia reacts with two moles of sodium hydroxide to produce two moles of aqueous sodium chloride solution and one mole of solid magnesium hydroxide.

  • 2H 2 O + 2Na → 2NaOH + H 2

2 moles of water vapour react with 2 moles of sodium metal, yielding 2 moles of solid sodium hydroxide and 1 mol of hydrogen gas.

Practise Questions on Balanced Chemical Equations

Balance the following equations-

1. (NH 4 ) 2 Cr 2 O 7 (s) → Cr 2 O 3 (s) + N 2 (g) + H 2 O(g)

2. Ca(OH) 2 + H 3 PO 4 → Ca 3 (PO 4 ) 2 + H 2 O

3. FeCl 3 + NH 4 OH → Fe(OH) 3 + NH 4 Cl

4. Al 2 (CO 3 ) 3 + H 3 PO 4 → AlPO 4 + CO 2 + H 2 O

5. S 8 + F 2 → SF 6

Click the PDF to check the answers for Practice Questions. Download PDF

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Balancing Chemical Equations with Practice Problems

July 16, 2019 By Leah4sci Leave a Comment

Balancing Chemical Equations Stoichiometry Series by Leah4sci

T he videos below will show you a simple yet efficient approach to make sure you quickly account for all your numbers (minimal guessing) and don’t lose track of your work along the way.

Balancing Equations Video 1 – Introduction to Balancing with practice problems

Let’s start with the simple approach to balancing equations including practice problems so that you get comfortable with the process.

(Watch on  YouTube : Balancing Equations . Click CC for transcription.)

Balancing Combustion Reaction Equations – Balancing Video 2

Balancing combustion reactions can be tricky due to having oxygen appear in both products. Many professors solve this with a fraction but you can’t have ‘half’ oxygens in a balanced equation. Learn a simpler method for balancing combustion reactions while avoiding fractions.

(Watch on  YouTube : Balancing Combustion . Click CC for transcription.)

Balancing Acid Base Reaction Equations – Balancing Video 3

Acid Base and Ionic reactions can be balanced the same way shown above, with an additional time-saving shortcut to help you deal with all the atoms within each polyatomic ion as taught in the video below.

(Watch on  YouTube : Balancing Acid Base . Click CC for transcription.)

<– Watch Previous Video:  Chemical Reactions –> Watch Next Video:  Coming Soon

This is Video 12 in the Stoichiometry & Reactions Video Series.  Click HERE for the entire series .

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Alkene Reactions Overview Cheat Sheet – Organic Chemistry

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KET Keto enol tautomerization reaction and mechanism leah4sci

Keto Enol Tautomerization Reaction and Mechanism

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  4. Chemistry Practice Problems: Balancing Chemical Equations

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  6. Chemistry Worksheet: Easy Balancing Equations Problems by Ray Byle

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COMMENTS

  1. Balancing chemical equations 1 (practice)

    Mg (OH) 2 + HCl → MgCl 2 + H 2 O Note: All reactants and products require a coefficient of at least one. Stuck? Review related articles/videos or use a hint. Report a problem Do 4 problems Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more.

  2. 5.1.1: Practice Problems- Writing and Balancing Chemical Equations

    PROBLEM 5.1.1.7 5.1.1. 7. A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process. The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide. The second step is the formation of ...

  3. Balancing Chemical Equations: Practice and Review

    The ultimate goal for balancing chemical equations is to make both sides of the reaction, the reactants and the products, equal in the number of atoms per element. This stems from the universal law of the conservation of mass, which states that matter can neither be created nor destroyed.

  4. Balancing Chemical Equations Practice Sheet

    Balancing chemical equations requires practice. Once you've done it a few times, it becomes easier and easier. This balancing chemical equations practice sheet has ten more unbalanced chemical equations to solve. Download a PDF of this worksheet here. A PDF of the answer key is also available here.

  5. Balancing, Identifying & Predicting Chemical Equations Quiz

    This online quiz is intended to give you extra practice in balancing, identifying and predicting a random selection of over 150 chemical equations. This quiz aligns with the following NGSS standard (s): HS-PS1-2, HS-PS1-7 Select your preferences below and click 'Start' to give it a try!

  6. PDF Balancing Equations: Practice Problems

    Balancing Equations: Practice Problems 1. Balance each of the following equations. Balancing Equations: Answers to Practice Problems Balanced equations. (Coefficients equal to one (1) do not need to be shown in your answers). 2 Fe+ 3 Cl2 − −→ 2 FeCl3 4 Fe+ 3 O2 − − → 2 Fe2O3 2 FeBr + − −→ + 3 3 H2SO4 1 Fe2(SO4) 3 (d) 1 C4H6O3 + 1 H2O − −→ 2 C2H4O2

  7. Balancing Equations Practice Quiz

    Science, Tech, Math › Science Balancing Equations Practice Quiz Here are 10 unbalanced equations. Select the correct balanced equation. Take this online quiz to practice balancing chemical equations. Vladimir Nenov / EyeEm / Getty Images By Anne Marie Helmenstine, Ph.D. Updated on October 26, 2020 1. ___ SnO₂ + ___ H₂ → ___ Sn + ___ H₂O

  8. ‪Balancing Chemical Equations‬

    Learn how to balance chemical equations by using the law of conservation of mass and the coefficients of reactants and products. Practice with different levels of difficulty and get immediate feedback. Compare your results with real-life examples and simulations of chemical reactions.

  9. Balancing equations (apply) (practice)

    Visually understanding balancing chemical equations. Balancing another combustion reaction. Apply: balancing equations. Science > High school chemistry ... You might need: Calculator. Problem. The following equation represents the reaction between solid aluminum and hydrochloric acid. Balance the equation by filling in the correct coefficients ...

  10. Balancing simple chemical equations (practice)

    Balance the following chemical equation: Note: Do not leave any boxes empty. Fill all the boxes with natural numbers. H 2 + O 2 → H 2 O. Stuck? Use a hint. Report a problem. Do 4 problems. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more.

  11. 4.1 Writing and Balancing Chemical Equations

    The chemical equation described in section 4.1 is balanced, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of ...

  12. Balancing Chemical Equations

    How do you know if a chemical equation is balanced? What can you change to balance an equation? Play a game to test your ideas!

  13. ChemTeam: Balancing Chemical Equations: Problems #1

    Balancing Chemical Equations Problems #1 - 10. Thirty examples Problems 11-25 Problems 26-45 Problems 46-65; Six "balancing by groups" problems ... The chemical equation is balanced in a chemically-correct sense with the fractional coefficients. Problem #9: C 5 H 11 NH 2 + O 2---> CO 2 + H 2 O + NO 2. Solution: 1) Balance the hydrogens first:

  14. Balancing Chemical Equations (Simplified) Practice Problems

    16 PRACTICE PROBLEM. Consider the following chemical reactions with compounds written in words instead of chemical formulas. Write balanced equations for these reactions using chemical formulas and all the physical states. A) Sodium + chlorine → Sodium chloride. B) Calcium hydride + hydrochloric acid → calcium chloride + hydrogen.

  15. Balancing Chemical Equations Practice Problems

    0:00 / 14:56 Introduction to Balancing Chemical Equations Equation balancing will make sense! Here, we will do a bunch of practice problems for balancing chemical equations....

  16. Balancing Chemical Equations Video Tutorial & Practice

    1 concept Balancing Chemical Equations 1m 0 Comments Mark as completed Was this helpful? 41 2 example Balancing Chemical Equations Example 1 3m 4 Comments Mark as completed Was this helpful? 22 3 Problem Write the balanced equation for the following by inserting the correct coefficients in the blanks. 2m 1 Comment Mark as completed

  17. Balancing chemical equations practice problems Flashcards

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  18. Balancing Chemical Equations Step by Step Practice Problems

    0:00 / 6:22 Balancing Chemical Equations Step by Step Practice Problems | How to Pass Chemistry Melissa Maribel 343K subscribers Subscribe Subscribed 552K views 6 years ago Fundamental Topics...

  19. Balancing Chemical Equations Video Tutorial & Practice

    Balancing Chemical Equations Practice Problems. Tyler DeWitt. 536. views. 1. rank. 08:54. How to Balance a Chemical Equation EASY. The Science Classroom. 416. views. 06:23. ... A key step in balancing chemical equations is correctly identifying the formulas of the reactants and products. For example, consider the reaction between calcium oxide ...

  20. Balanced Chemical Equations Questions

    Q1. A balanced chemical equation is in accordance with- Multiple proportion Reciprocal proportion Conservation of mass Definite proportions Correct Answer: (c) Law of Conservation of Mass Q2. The correct balanced equation for the reaction __C2H6O + __O2 → __CO2 + __H2O is- 2C 2 H 6 O + O 2 → CO 2 + H 2 O C 2 H 6 O + 3O 2 → 2CO 2 + 3H 2 O

  21. Balancing Chemical Equations Practice Problems

    Balancing Chemical Equations Practice Problems 44 problems 1 PRACTICE PROBLEM Provide a balanced chemical equation for the production of the II-IV semiconductor CdS from the reaction of dimethylcadmium [Cd (CH 3) 2] and diethyl sulfide [ (C 2 H 5) 2 S]. The byproducts of the reaction are ethane (C 2 H 6) and butane (C 4 H 10 ). 22 5

  22. Balancing Chemical Equations with Practice Problems

    Balancing Acid Base Reaction Equations - Balancing Video 3. Acid Base and Ionic reactions can be balanced the same way shown above, with an additional time-saving shortcut to help you deal with all the atoms within each polyatomic ion as taught in the video below. (Watch on YouTube: Balancing Acid Base. Click CC for transcription.)