stoichiometry sample problems

Chemistry Steps

General Chemistry

Stoichiometry.

This is a comprehensive, end-of-chapter set of practice problems on stoichiometry that covers balancing chemical equations, mole-ratio calculations, limiting reactants, and percent yield concepts. 

The links to the corresponding topics are given below.

• The Mole and Molar Mass
• Molar Calculations
• Percent Composition and Empirical Formula
• Stoichiometry of Chemical Reactions

Limiting Reactant

• Reaction/Percent Yield
• Stoichiometry Practice Problems

Balance the following chemical equations:

a) HCl + O 2 → H 2 O + Cl 2

b) Al(NO 3 ) 3 + NaOH → Al(OH) 3 + NaNO 3

c) H 2 + N 2 → NH 3

d) PCl 5 + H 2 O → H 3 PO 4 + HCl

e) Fe + H 2 SO 4 → Fe 2 (SO 4 ) 3 + H 2

f) CaCl 2 + HNO 3 → Ca(NO 3 ) 2 + HCl

g) KO 2 + H 2 O → KOH + O 2 + H 2 O 2

h) Al + H 2 O → Al 2 O 3 + H 2

i) Fe + Br 2 → FeBr 3

j) Cu + HNO 3 → Cu(NO 3 ) 2 + NO 2 + H 2 O

k) Al(OH) 3 → Al 2 O 3 + H 2 O

l) NH 3 + O 2 → NO + H 2 O

m) Ca(AlO 2 ) 2 + HCl → AlCl 3 + CaCl 2 + H 2 O

n) C 5 H 12 + O 2 → CO 2 + H 2 O

o) P 4 O 10 + H 2 O → H 3 PO 4

p) Na 2 CrO 4 + Pb(NO 3 ) 2 → PbCrO 4 + NaNO 3

q) MgCl 2 + AgNO 3 → AgCl + Mg(NO 3 ) 2

r) KClO 3 → KClO 4 + KCl

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

Consider the balanced equation:

C 5 H 12 + 8 O 2 → 5CO 2 + 6H 2 O

Complete the table showing the appropriate number of moles of reactants and products.

How many grams of CO 2  and H 2 O are produced from the combustion of 220. g of propane (C 3 H 8 )?

C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g)

How many grams of CaCl 2 can be produced from 65.0 g of Ca(OH) 2 according to the following reaction,

Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O

How many moles of oxygen are formed when 75.0 g of Cu(NO 3 ) 2 decomposes according to the following reaction?

2Cu(NO 3 ) 2   → 2CuO + 4NO 2  + O 2

How many grams of MnCl 2  can be prepared from 52.1 grams of MnO 2 ?

MnO 2  + 4HCl → MnCl 2  + Cl 2  + 2H 2 O

Determine the mass of oxygen that is formed when an 18.3-g sample of potassium chlorate is decomposed according to the following equation:

2KClO 3 (s) → 2KCl(s) + 3O 2 (g).

How many grams of H 2 O will be formed when 48.0 grams H 2 are mixed with excess hydrogen gas?

2H 2  + O 2 → 2H 2 O

Consider the chlorination reaction of methane (CH4):

CH 4 (g) + 4Cl 2 (g) → CCl 4 (g) + 4HCl(g)

How many moles of CH 4 were used in the reaction if 51.9 g of CCl4 were obtained?

How many grams of Ba(NO 3 ) 2 can be produced by reacting 16.5 g of HNO 3 with an excess of Ba(OH) 2 ?

Ethanol can be obtained by fermentation – a complex chemical process breaking down glucose to ethanol and carbon dioxide.

                                                  C 6 H 12 O 6    →    2C 2 H 5 OH   +    2CO 2                                                       glucose                   ethanol

How many mL of ethanol (d =0.789 g/mL) can be obtained by this process starting with 286 g of glucose?

36.0 g of butane (C 4 H 10 ) was burned in an excess of oxygen and the resulting carbon dioxide (CO 2 ) was collected in a sealed vessel.

2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O

How many grams of LiOH will be necessary to consume all the CO 2 from the first reaction?

2LiOH + CO 2 → Li 2 CO 3 + H 2 O

13. Which statement about limiting reactant is correct?

a) The limiting reactant is the one in a smaller quantity.

b) The limiting reactant is the one in greater quantity.

c) The limiting reactant is the one producing less product.

d) The limiting reactant is the one producing more product.

Find the limiting reactant for each initial amount of reactants.

4NH 3 + 5O 2 → 4NO + 6H 2 O

a) 2 mol of NH 3 and 2 mol of O 2

b) 2 mol of NH 3 and 3 mol of O 2

c) 3 mol of NH 3 and 3 mol of O 2

d) 3 mol of NH 3 and 2 mol of O 2

Note:  This is not a multiple-choice question. Each row represents a separate question where you need to determine the limiting reactant.

How many g of hydrogen are left over in producing ammonia when 14.0 g of nitrogen is reacted with 8.0 g of hydrogen?

N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

How many grams of PCl 3 will be produced if 130.5 g Cl 2 is reacted with 56.4 g P 4 according to the following equation?

6Cl 2 (g) + P 4 (s) → 4PCl 3 (l)

How many grams of sulfur can be obtained if 12.6 g H 2 S is reacted with 14.6 g SO 2 according to the following equation?

2H 2 S(g) + SO 2 (g) → 3S(s) + 2H 2 O(g)

The following equation represents the combustion of octane, C 8 H 18 , a component of gasoline:

2C 8 H 18 (g) + 25O 2 (g) → 16CO 2 (g) + 18H 2 O(g)

Will 356 g of oxygen be enough for the complete combustion of 954 g of octane?

When 140.0 g of AgNO 3 was added to an aqueous solution of NaCl, 86.0 g of AgCl was collected as a white precipitate. Which salt was the limiting reactant in this reaction? How many grams of NaCl were present in the solution when AgNO 3 was added?

AgNO 3 (aq) + NaCl(aq) → AgCl(s) + NaNO 3 (aq)

Consider the reaction between MnO 2 and HCl:

MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O

What is the theoretical yield of MnCl 2 in grams when 165 g of MnO 2 is added to a solution containing 94.2 g of HCl?

Percent Yield

21. In a chemistry experiment, a student obtained 5.68 g of a product. What is the percent yield of the product if the theoretical yield was 7.12 g?

When 38.45 g CCl 4 is reacted with an excess of HF, 21.3 g CCl 2 F 2 is obtained. Calculate the theoretical and percent yields of this reaction.

CCl 4 + 2HF → CCl 2 F 2 + 2HCl

Iron(III) oxide reacts with carbon monoxide according to the equation:

Fe 2 O 3 ( s ) + 3CO( g ) → 2Fe( s ) + 3CO 2 ( g )

What is the percent yield of this reaction if 623 g of iron oxide produces 341 g of iron?

Determine the percent yield of the reaction if 77.0 g of CO 2  are formed from burning 2.00 moles of C 5 H 12  in 4.00 moles of O 2 .

C 5 H 12 + 8 O 2 → 5CO 2  + 6H 2 O

The percent yield for the following reaction was determined to be 84%:

N 2 ( g ) + 2H 2 ( g ) → N 2 H 4 ( l )

How many grams of hydrazine (N 2 H 4 ) can be produced when 38.36 g of nitrogen reacts with 6.68 g of hydrogen?

Silver metal can be prepared by reducing its nitrate, AgNO 3  with copper according to the following equation:

Cu( s ) + 2AgNO 3 ( aq ) → Cu(NO 3 ) 2 ( aq ) + 2Ag( s )

What is the percent yield of the reaction if 71.5 grams of Ag was obtained from 132.5 grams of AgNO 3  ?

Industrially, nitric acid is produced from ammonia by the Ostwald process in a series of reactions:

4NH 3 ( g ) + 5O 2 ( g ) → 4NO( g ) + 6H 2 O( l )

2NO( g ) + O 2 ( g ) → 2NO 2 ( g )

2NO 2 ( g ) + H 2 O( l ) → HNO 3 ( aq ) + HNO 2 ( aq )

Considering that each reaction has an 85% percent yield, how many grams of NH 3 must be used to produce 25.0 kg of HNO 3 by the above procedure?

Aspirin (acetylsalicylic acid) is widely used to treat pain, fever, and inflammation. It is produced from the reaction of salicylic acid with acetic anhydride. The chemical equation for aspirin synthesis is shown below:

In one container, 10.00 kg of salicylic acid is mixed with 10.00 kg of acetic anhydride.

a)  Which reactant is limiting? Which is in excess? b)  What mass of excess reactant is left over? c)  What mass of aspirin is formed assuming 100% yield (Theoretical yield)? d)  What mass of aspirin is formed if the reaction yield is 70.0% ? e)  If the actual yield of aspirin is 11.2 kg, what is the percent yield? f)  How many kg of salicylic acid is needed to produce 20.0 kg of aspirin if the reaction yield is 85.0% ?

3 thoughts on “Stoichiometry Practice Problems”

You forgot the subscript 3 for O in the molecular formula for acetic anhydride and the reaction is not balanced as written. For part F) it’s 18.1 kg and not1.81 kg as written in the final line of the solution.

Thanks for letting me know! Fixed.

You’re welcome!

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What is Stoichiometry? Examples and Practice

• The Albert Team
• Last Updated On: March 17, 2024

Have you ever wondered how chemists know exactly how much of each substance to use in a reaction? The answer lies in a fundamental concept called stoichiometry. This crucial aspect of chemistry helps scientists and students alike understand the quantitative relationships in chemical reactions, ensuring that no atom is wasted. In this post, we’ll delve into what stoichiometry is, explore some engaging stoichiometry examples, and provide you with practical stoichiometry practice problems. Whether you’re just getting started or looking to sharpen your skills, this guide will equip you with the knowledge and tools to tackle stoichiometry problems with confidence.

What We Review

What is Stoichiometry?

Stoichiometry is like the recipe for a cake, but instead of baking, we’re dealing with chemical reactions. At its core, stoichiometry is the study of the quantitative relationships between the reactants and products in chemical reactions. When chemists conduct experiments, they need to know how much of each reactant to use and what amount of product to expect. Stoichiometry provides these answers, ensuring that chemical reactions are efficient and effective.

Understanding the Basics

To grasp stoichiometry, you first need to be familiar with a few key concepts:

• Moles: Just like a dozen represents 12 items, a mole is a specific number of molecules (approximately 6.022\times10^{23} molecules, known as Avogadro’s number).
• Molar Mass: This is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). It tells you how much one mole of a substance weighs.
• Mole Ratios: These ratios come from the coefficients of a balanced chemical equation. They tell you the proportion of reactants and products in a reaction.

Real-World Applications

Stoichiometry isn’t just a topic in your chemistry textbook; it has real-world applications. Pharmacists use stoichiometry to mix medications, environmental scientists use it to track pollutant levels, and engineers apply it to design reactors. Understanding stoichiometry means you’re learning a skill that scientists and professionals use daily to make the world work.

Stoichiometry Examples

Understanding stoichiometry becomes clearer with practical examples. Let’s explore a couple of scenarios to illustrate how stoichiometry helps us solve chemical equations.

Example 1: Water Formation

Consider the chemical reaction where hydrogen and oxygen combine to form water:

If we start with 36.0 grams of H_2 and have an excess of O_2 , how many grams of H_2O can we produce? The molar mass of H_2 is 2.02 g/mol, and for H_2O , it’s 18.02 g/mol.

First, convert the mass of H_2 to moles:

Applying the mole ratio from the balanced equation, calculate the moles of H_2O produced:

Then, convert moles of H_2O to grams:

Thus, 36.0 grams of H_2 can produce 321.15 grams of H_2O .

Example 2: Ammonia Synthesis

Now, let’s examine another one of our stoichiometry examples. This problem examines the synthesis of ammonia ( NH_3 ) from nitrogen ( N_2 ) and hydrogen ( H_2 ) in a reaction conducted at STP (standard temperature and pressure):

If we begin with 22.4 liters of N_2 (1 mole at STP) and have an excess of H_2 , how many grams of NH_3 can be produced? The molar mass of NH_3 is 17.03 g/mol.

Using the mole ratio from the chemical equation:

Then, convert moles of NH_3 to grams:

This result shows that 22.4 liters of N_2 at STP can produce 34.06 grams of NH_3 .

Guided Stoichiometry Practice

To master stoichiometry, practice is key. In this section, we’ll explore strategies to approach stoichiometry problems and provide some guided practice examples. These strategies will help you develop a systematic approach to solving stoichiometry questions, enhancing your problem-solving skills.

Understanding the Approach

Identify What’s Given and What’s Unknown: Start by determining what information is provided in the problem and what you need to find. This could be the amount of reactants, products, or even the coefficients in a balanced chemical equation.

Write the Balanced Chemical Equation: Ensure you have the correct balanced chemical equation for the reaction you’re analyzing. This step is crucial as it sets the foundation for the stoichiometric calculations.

Use Mole Ratios: Utilize the coefficients from the balanced equation to establish mole ratios. These ratios are the heart of stoichiometry, allowing you to convert between amounts of reactants and products.

Perform the Calculations: With the mole ratios, you can perform calculations to find the unknown quantities. Remember to keep track of your units, ensuring they are consistent throughout the calculation.

Guided Example

Let’s practice with a guided example. Consider the combustion of propane ( C_3H_8 ) in oxygen to produce carbon dioxide and water:

Suppose you start with 44.1 grams of propane ( C_3H_8 ) and have an excess of oxygen ( O_2 ). How many grams of carbon dioxide ( CO_2 ) are produced?

Given: 44.1 grams of C_3H_8 The molar mass of C_3H_8 is 44.1 g/mol, and for CO_2 , it’s 44.0 g/mol.

Unknown: Grams of CO_2 produced

First, convert the mass of C_3H_8 to moles:

Mole Ratio: From the balanced equation, the mole ratio of C_3H_8 to CO_2 is 1:3.

Calculation:

Now, convert the moles of CO_2 to grams:

So, 44.1 grams of C_3H_8 produces 132.0 grams of CO_2 .

Stoichiometry Practice Problems

Now that you’ve seen stoichiometry in action through examples, it’s time to test your knowledge with some practice problems. Try to solve these problems on your own before checking the solutions provided. This will help you understand stoichiometry better and prepare you for similar problems in the future.

When magnesium burns in the air, it reacts with oxygen to form magnesium oxide. The balanced chemical equation for this reaction is:

If you start with 24.0 grams of magnesium ( Mg ) and have an excess of oxygen ( O_2 ), how many grams of magnesium oxide ( MgO ) will be produced? The molar mass of Mg is 24.3 g/mol, and MgO is 40.3 g/mol.

Hydrochloric acid reacts with sodium hydroxide to produce sodium chloride and water in the following reaction:

If 36.5 grams of hydrochloric acid ( HCl ) react with sufficient sodium hydroxide ( NaOH ), how many grams of sodium chloride ( NaCl ) are produced? Assume the molar mass of HCl is 36.5 g/mol and NaCl is 58.4 g/mol.

Nitrogen gas can react with hydrogen gas to form ammonia through the following reaction:

Calculate the amount of ammonia ( NH_3 ) produced (in grams) when 28.0 liters of nitrogen gas ( N_2 ) react with an excess of hydrogen gas ( H_2 ) at STP. The molar mass of NH_3 ​ is [/latex]17.03[/latex] g/mol.

Stoichiometry Practice Problems Solutions

Here, we’ll go through the solutions to the stoichiometry practice problems we presented earlier. These solutions will help you understand the step-by-step process involved in solving stoichiometry problems.

Solution to Problem 1:

For the reaction 2Mg + O_2 \rightarrow 2MgO , we’re starting with 24.0 grams of Mg . First, convert the mass of Mg to moles:

Using the stoichiometry of the reaction, convert moles of Mg to moles of MgO (the ratio is 1:1):

Now, convert moles of MgO to grams:

So, 24.0 grams of Mg will produce 39.81 grams of MgO .

Solution to Problem 2:

For the reaction HCl + NaOH \rightarrow NaCl + H_2O , we start with 36.5 grams of HCl . Convert the mass of HCl to moles:

The mole ratio of HCl to NaCl is 1:1, so:

Now, convert moles of NaCl to grams:

Thus, 36.5 grams of HCl will produce 58.44 grams of NaCl .

Solution to Problem 3:

For the reaction N_2 + 3H_2 \rightarrow 2NH_3 , starting with 28.0 liters of N_2 ​ at STP (which corresponds to 1.00 mole of N_2 ​):

Using the stoichiometry of the reaction (1:2 ratio of N_2 to NH_3 ):

Convert moles of NH_3 ​ to grams:

So, 28.0 liters of N_2 ​ will produce 42.575 grams of NH_3 ​.

Stoichiometry is a cornerstone concept in chemistry that enables us to predict and quantify the outcomes of chemical reactions. By understanding the relationships between reactants and products, chemists can conduct reactions efficiently and effectively. Throughout this post, we’ve explored what stoichiometry is, provided clear examples, and offered practice problems to help you build your skills.

Remember, the key to mastering stoichiometry lies in practice. The more you work through problems, the more intuitive these concepts will become. Whether you’re balancing equations, calculating molar masses, or determining the amounts of reactants and products, each step you take strengthens your understanding of chemistry.

We encourage you to revisit these examples and practice problems regularly, and don’t hesitate to explore more complex scenarios as you become more comfortable with the basics. Stoichiometry is not just a topic for the classroom; it’s a tool that scientists use to make real-world decisions in industries, research, and environmental science.

So, keep practicing, stay curious, and remember that every chemical equation tells a story of transformation and interaction. With stoichiometry, you’re equipped to understand and narrate these fascinating stories of science.

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Solving Stoichiometry Problems

Core Concepts

In this tutorial, you will learn what is stoichiometry and the different types of problems involving it. You will go through several examples to practice and master the content!

Topics Covered in Other Articles

Balancing chemical equations.

• What is a Chemical Reaction?
• Calculating Molar Mass
• Percent Yield Calculation
• How to Calculate Molarity
• What Are Significant Figures

Stoichiometry Definition

What is stoichiometry.

Stoichiometry is math having to do with chemical reactions. There are different types of calculations you can perform; stoichiometry with moles is the most common, but you can also do math with masses and even percentages. Read about the origins of stoichiometry here ! Learn what is mole in chemistry .

Stoichiometric Ratio

A stoichiometric ratio comes into play when talking about the relationships of elements or molecules in specific problems. This is the exact ratio between the coefficients of the reactants and products needed for a reaction to proceed normally. Let’s work through some problems you may see when learning about stoichiometry.

Stoichiometry Problems

A very common type of stoichiometric problem is balancing equations. This is an important chemistry skill to have because you have to have the correct ratio of reactants and products in order for a reaction to proceed; this is also an important foundation for organic chemistry. Although we have a tutorial on balancing equations , let’s look at one example.

Balance the following reaction:

The main idea when balancing equations is that there should be the same number of each element on both sides of the reaction. You can balance the carbons and the hydrogens first, then move onto the oxygen. The balanced equation should look like this:

Example – Using Stoichiometric Ratio (Moles)

Use the equation below to solve the problem.

Example – Using Stoichiometric Ratio (Mass)

Let’s use the same equation as the problem above.

Similar to the previous problem, by using the stoichiometric ratio of reactant to product, you can find the answer. Dimensional analysis is used to go from grams of C 2 H 5 OH to molar mass to mole (stoichiometric) ratio and back to grams.

Stoichiometry Practice Problems

CaCl 2 and AgNO 3 react according to the following equation:

Hexane combusts according to the following reaction:

Stoichiometry Practice Problem Solutions

For more help, view our interactive lecture on introducing stoichiometry problems!

And this video explaining how to solve reaction stoichiometry problems, further reading.

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Chemistry: Stoichiometry of Reactions

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5.3: Stoichiometry Calculations

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Learning Objectives

• To balance equations that describe reactions in solution.
• To calculate the quantities of compounds produced or consumed in a chemical reaction.
• To solve quantitative problems involving the stoichiometry of reactions in solution.

A balanced chemical equation gives the identity of the reactants and the products as well as the accurate number of molecules or moles of each that are consumed or produced. Stoichiometry is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. A stoichiometric quantity is the amount of product or reactant specified by the coefficients in a balanced chemical equation. This section describes how to use the stoichiometry of a reaction to answer questions like the following: How much oxygen is needed to ensure complete combustion of a given amount of isooctane? (This information is crucial to the design of nonpolluting and efficient automobile engines.) How many grams of pure gold can be obtained from a ton of low-grade gold ore? (The answer determines whether the ore deposit is worth mining.) If an industrial plant must produce a certain number of tons of sulfuric acid per week, how much elemental sulfur must arrive by rail each week?

All these questions can be answered using the concepts of the mole, molar and formula masses, and solution concentrations, along with the coefficients in the appropriate balanced chemical equation.

Stoichiometry Problems

When carrying out a reaction in either an industrial setting or a laboratory, it is easier to work with masses of substances than with the numbers of molecules or moles. The general method for converting from the mass of any reactant or product to the mass of any other reactant or product using a balanced chemical equation is outlined in and described in the following text.

Steps in Converting between Masses of Reactant and Product

• Convert the mass of one substance (substance A) to the corresponding number of moles using its molar mass.
• From the balanced chemical equation, obtain the number of moles of another substance (B) from the number of moles of substance A using the appropriate mole ratio (the ratio of their coefficients).
• Convert the number of moles of substance B to mass using its molar mass. It is important to remember that some species are present in excess by virtue of the reaction conditions. For example, if a substance reacts with the oxygen in air, then oxygen is in obvious (but unstated) excess.

Converting amounts of substances to moles—and vice versa—is the key to all stoichiometry problems, whether the amounts are given in units of mass (grams or kilograms), weight (pounds or tons), or volume (liters or gallons).

To illustrate this procedure, consider the combustion of glucose. Glucose reacts with oxygen to produce carbon dioxide and water:

\[ C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \label{3.6.1} \]

Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain does not run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam, how many grams of carbon dioxide will you produce and exhale into the exam room?

The initial step in solving a problem of this type is to write the balanced chemical equation for the reaction. Inspection shows that it is balanced as written, so the strategy outlined above can be adapted as follows:

1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:

\[ moles \, glucose = 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose } = 0.251 \, mol \, glucose \nonumber \]

2. According to the balanced chemical equation, 6 mol of CO 2 is produced per mole of glucose; the mole ratio of CO 2 to glucose is therefore 6:1. The number of moles of CO 2 produced is thus

\[ moles \, CO_2 = mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber \]

\[ = 0.251 \, mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber \]

\[ = 1.51 \, mol \, CO_2 \nonumber \]

3. Use the molar mass of CO 2 (44.010 g/mol) to calculate the mass of CO 2 corresponding to 1.51 mol of CO 2 :

\[ mass\, of\, CO_2 = 1.51 \, mol \, CO_2 \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.5 \, g \, CO_2 \nonumber \]

These operations can be summarized as follows:

\[ 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose} \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose} \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.4 \, g \, CO_2 \nonumber \]

Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (Remember that you should generally carry extra significant digits through a multistep calculation to the end to avoid this!) This amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. Similar methods can be used to calculate the amount of oxygen consumed or the amount of water produced.

The balanced chemical equation was used to calculate the mass of product that is formed from a certain amount of reactant. It can also be used to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example \(\PageIndex{1}\), the mass of one reactant that is required to consume a given mass of another reactant.

Example \(\PageIndex{1}\): The US Space Shuttle

The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb).

The US space shuttle Discovery during liftoff . The large cylinder in the middle contains the oxygen and hydrogen that fueled the shuttle’s main engine .

Given : reactants, products, and mass of one reactant

Asked for : mass of other reactant

• Write the balanced chemical equation for the reaction.
• Convert mass of oxygen to moles. From the mole ratio in the balanced chemical equation, determine the number of moles of hydrogen required. Then convert the moles of hydrogen to the equivalent mass in tons.

We use the same general strategy for solving stoichiometric calculations as in the preceding example. Because the amount of oxygen is given in tons rather than grams, however, we also need to convert tons to units of mass in grams. Another conversion is needed at the end to report the final answer in tons.

A We first use the information given to write a balanced chemical equation. Because we know the identity of both the reactants and the product, we can write the reaction as follows:

\[ H_2 (g) + O_2 (g) \rightarrow H_2O (g) \nonumber \]

This equation is not balanced because there are two oxygen atoms on the left side and only one on the right. Assigning a coefficient of 2 to both H 2 O and H 2 gives the balanced chemical equation:

\[ 2 H_2 (g) + O_2 (g) \rightarrow 2 H_2O (g) \nonumber \]

Thus 2 mol of H 2 react with 1 mol of O 2 to produce 2 mol of H 2 O.

1. B To convert tons of oxygen to units of mass in grams, we multiply by the appropriate conversion factors:

\[ mass \, of \, O_2 = 1.00 \, tn \times { 2000 \, lb \over tn} \times {453.6 \, g \over lb} = 9.07 \times 10^5 \, g \, O_2 \nonumber \]

Using the molar mass of O 2 (32.00 g/mol, to four significant figures), we can calculate the number of moles of O 2 contained in this mass of O 2 :

\[ mol \, O_2 = 9.07 \times 10^5 \, g \, O_2 \times {1 \, mol \, O_2 \over 32.00 \, g \, O_2} = 2.83 \times 10^4 \, mol \, O_2 \nonumber \]

2. Now use the coefficients in the balanced chemical equation to obtain the number of moles of H 2 needed to react with this number of moles of O 2 :

\[ mol \, H_2 = mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} \nonumber \]

\[ = 2.83 \times 10^4 \, mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} = 5.66 \times 10^4 \, mol \, H_2 \nonumber \]

3. The molar mass of H 2 (2.016 g/mol) allows us to calculate the corresponding mass of H 2 :

\[mass \, of \, H_2 = 5.66 \times 10^4 \, mol \, H_2 \times {2.016 \, g \, H_2 \over mol \, H_2} = 1.14 \times 10^5 \, g \, H_2 \nonumber \]

Finally, convert the mass of H2 to the desired units (tons) by using the appropriate conversion factors:

\[ tons \, H_2 = 1.14 \times 10^5 \, g \, H_2 \times {1 \, lb \over 453.6 \, g} \times {1 \, tn \over 2000 \, lb} = 0.126 \, tn \, H_2 \nonumber \]

The space shuttle had to be designed to carry 0.126 tn of H 2 for each 1.00 tn of O 2 . Even though 2 mol of H 2 are needed to react with each mole of O 2 , the molar mass of H 2 is so much smaller than that of O 2 that only a relatively small mass of H 2 is needed compared to the mass of O 2 .

Exercise \(\PageIndex{1}\): Roasting Cinnabar

Cinnabar, (or Cinnabarite) \(HgS\) is the common ore of mercury. Because of its mercury content, cinnabar can be toxic to human beings; however, because of its red color, it has also been used since ancient times as a pigment.

Alchemists produced elemental mercury by roasting cinnabar ore in air:

\[ HgS (s) + O_2 (g) \rightarrow Hg (l) + SO_2 (g) \nonumber \]

The volatility and toxicity of mercury make this a hazardous procedure, which likely shortened the life span of many alchemists. Given 100 g of cinnabar, how much elemental mercury can be produced from this reaction?

Calculating Moles from Volume

Quantitative calculations involving reactions in solution are carried out with masses , however, volumes of solutions of known concentration are used to determine the number of moles of reactants. Whether dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation give the number of moles of each reactant needed and the number of moles of each product that can be produced. An expanded version of the flowchart for stoichiometric calculations is shown in Figure \(\PageIndex{2}\). The balanced chemical equation for the reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used to determine the amounts of other species, as illustrated in the following examples.

The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations.

Example \(\PageIndex{2}\) : Extraction of Gold

Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN) 2 ] − ion. Gold is then recovered by reduction with metallic zinc according to the following equation:

\[ Zn(s) + 2[Au(CN)_2]^-(aq) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s) \nonumber \]

What mass of gold can be recovered from 400.0 L of a 3.30 × 10 −4 M solution of [Au(CN) 2 ] − ?

Given: chemical equation and molarity and volume of reactant

Asked for: mass of product

• Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN) 2 ] − present by multiplying the volume of the solution by its concentration.
• From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass.

A The equation is balanced as written; proceed to the stoichiometric calculation. Figure \(\PageIndex{2}\) is adapted for this particular problem as follows:

As indicated in the strategy, start by calculating the number of moles of [Au(CN) 2 ] − present in the solution from the volume and concentration of the [Au(CN) 2 ] − solution:

\( \begin{align} moles\: [Au(CN)_2 ]^- & = V_L M_{mol/L} \\ & = 400 .0\: \cancel{L} \left( \dfrac{3 .30 \times 10^{4-}\: mol\: [Au(CN)_2 ]^-} {1\: \cancel{L}} \right) = 0 .132\: mol\: [Au(CN)_2 ]^- \end{align} \)

B Because the coefficients of gold and the [Au(CN) 2 ] − ion are the same in the balanced chemical equation, assuming that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN) 2 ] − (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so the number of moles of gold must be converted to the corresponding mass using the molar mass of gold:

\( \begin{align} mass\: of\: Au &= (moles\: Au)(molar\: mass\: Au) \\ &= 0 .132\: \cancel{mol\: Au} \left( \dfrac{196 .97\: g\: Au} {1\: \cancel{mol\: Au}} \right) = 26 .0\: g\: Au \end{align}\)

At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth $1170.

\( 26 .0\: \cancel{g\: Au} \times \dfrac{1\: \cancel{troy\: oz}} {31 .10\: \cancel{g}} \times \dfrac{\$1400} {1\: \cancel{troy\: oz\: Au}} = \$1170 \)

Exercise \(\PageIndex{2}\) : Lanthanum Oxalate

What mass of solid lanthanum(III) oxalate nonahydrate [La 2 (C 2 O 4 ) 3 •9H 2 O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl 3 by adding a stoichiometric amount of sodium oxalate?

Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors, YouTube(opens in new window) [youtu.be]

Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation. Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced.

Stoichiometry & Limiting Reagents Quiz

This online quiz is intended to give you extra practice in performing stoichiometric conversions, including limiting reagent and percent yield problems. This quiz aligns with the following NGSS standard(s): HS-PS1-7

Select your preferences below and click 'Start' to give it a try!

H 2 + O 2 ---> H 2 O

2H 2 + O 2 ---> 2H 2 O

54.0 g / 32.0 g/mol = 1.6875 mol of O 2 Note the use of 32.0 and not 16.0. The chemical substance is O 2 . Students have been known to sometimes forget to write the subscript of 2 on a diatomic element (H 2 , N 2 , O 2 , F 2 , Cl 2 , Br 2 , I 2 )

H 2 ––– O 2

2 mol H 2 ––––––– 1 mol O 2

x ––––––––––– 1.6875 mol O 2 No unit is attached to the unknown 'x.' Note also that I did not round off. I'll do that at the end.

2 mol H 2   x ––––––– = ––––––––––– 1 mol O 2   1.6875 mol O 2 x = 3.375 mol of H 2 required

(3.375 mol) (2.016 g/mol) = 6.80 g (to three sig figs)

54.0 g O 2   1 mol O 2   2 mol H 2   2.016 g H 2   –––––––  x  –––––––  x  –––  x  –––––––  = 6.80 g H 2     32.0 g O 2   1 mol O 2   1 mol H 2   When doing DA, it is very helpful to identify each unit as being one chemical or the other. This will help cut down on confusion when you are placing values, deciding if a given value goes in the numerator or the denominator.

105.0 g / 18.015 g/mol = 5.82848 mol of H 2 O I rounded off some, but I made sure to keep more digits than what I will round off to at the end.

H 2 –––– H 2 O

2 mol H 2 ––––––––– 2 mol H 2 O

x ––––––––––––––– 5.82848 mol H 2 O

2 mol H 2   x ––––––––– = –––––––––––––– 2 mol H 2 O   5.82848 mol H 2 O x = 5.82848 mol of H 2 required

(5.82848 mol) (2.016 g/mol) = 11.75 g of H 2 (to four sig figs)

105.0 g H 2 O   1 mol H 2 O   2 mol H 2   2.016 g H 2   ––––––––––  x  –––––––––––  x  ––––––––––  x  –––––––  = 11.75 g H 2     18.015 g H 2 O   2 mol H 2 O   1 mol H 2   Remember, in dimensional analysis, you cancel units on the diagonal, never up and down. This is why you write the substance in the unit. 'mol H 2 ' and 'mol H 2 O' do not cancel.

N 2 + H 2 ---> NH 3

N 2 + 3H 2 ---> 2NH 3

85.2 g / 17.0307 g/mol = 5.00273 mol

H 2 –––– NH 3

3   x ––– = ––––––– 2   5.00273 x = 7.504095 mol of H 2

(7.504095 mol) (2.016 g/mol) = 15.1 g (to three sig figs)

85.2 g NH 3   1 mol NH 3   3 mol H 2   2.016 g H 2   –––––––––  x  ––––––––––––  x  –––––––––  x  –––––––––  = 15.1 g H 2     17.0307 g NH 3   2 mol NH 3   1 mol H 2  

AuCl 3 ---> Au + Cl 2

2AuCl 3 ---> 2Au + 3Cl 2

Let x = the moles of AuCl 3     64.00 g x = ––––––––––––     303.32 g/mol x = 0.210998 mol of AuCl 3 The ChemTeam has heard many variations of this: "But how did you know to convert grams of AuCl 3 to moles?"

AuCl 3 ––––– Cl 2

2 mol AuCl 3   0.210998 mol AuCl 3 –––––––––– = –––––––––––––––– 3 mol Cl 2   x x = 0.316497 mol of Cl 2

(0.316497 mol) (70.906 g/mol) = 22.44 g (to four sig figs)

64.00 g AuCl 3   1 mol AuCl 3   3 mol Cl 2   70.906 g Cl 2   –––––––––  x  ––––––––––––  x  –––––––––  x  –––––––––  = 22.44 g Cl 2     303.32 g AuCl 3   2 mol AuCl 3   1 mol Cl 2  

3AgNO 3 + AlCl 3 ---> 3AgCl + Al(NO 3 ) 3

200. g     –––––––––––– = 1.499914 mol of AlCl 3 133.341 g/mol     I picked AlCl 3 because it was the substance has a gram amount associated with it in the problem.

AgCl ––––– AlCl 3 3 mol AgCl   x ––––––––– = –––––––––––– 1 mol AlCl 3   1.499914 mol AlCl 3 x = 4.499742 mol of AgCl The 'x' in the right-hand ratio is associated with the substance we are trying to calculate an amount for (the AgCl). Look for phrases like "Calculate the mass of . . ." or "Determine the mass of . . . " in the problem statement.

(4.499742 mol) (143.323 g/mol) = 645 g (to three sig figs)

1 mol AlCl 3   1.499914 mol AlCl 3 –––––––––– = ––––––––––––––– 3 mol AgCl   x

200. g AuCl 3   1 mol AlCl 3   3 mol AgCl   143.323 g AgCl   –––––––––––  x  ––––––––––––  x  –––––––––  x  ––––––––––––  = 645 g AgCl     133.341 g AlCl 3   1 mol AlCl 3   1 mol AgCl  

2KI + Pb(NO 3 ) 2 ---> PbI 2 + 2KNO 3

30.0 g     –––––––––––– = 0.180725 mol of KI 165.998 g/mol    

This ratio: 2 ––– 1

0.180725 –––––––– x

2   0.180725 –– = –––––––– 1   x

(0.0903625 mol) (461.01 g/mol) = 41.6 g (to three sig figs)

Al(NO 3 ) 3 + Mg ---> Mg(NO 3 ) 2 + Al

2Al(NO 3 ) 3 + 3Mg ---> 3Mg(NO 3 ) 2 + 2Al

92.0 g     –––––––––– = 3.4099 mol of Al 26.98 g/mol    

Al –––––––– Al(NO 3 ) 3

2   3.4099 –– = ––––––– 2   x x = 3.4099 mol 3 ) 3 , NOT moles of Al

(3.4099 mol) (212.994 g/mol) = 726 g (to three sig figs)

2Au + 3Cl 2 ---> 2AuCl 3

100.0 g     –––––––––– = 1.41032 mol of Cl 2 70.906 g/mol    

3   1.41032 mol –– = ––––––––––– 2   x x = 0.940213 mol

(0.940213 mol) (303.329 g/mol) = 285 g

volume of Al foil ---> (1.00 cm) (1.00 cm) (0.0540 cm) = 0.0540 cm 3 Note the change of mm to cm.

Note the use of the density of aluminum.

2Al + 3Br 2 ---> 2AlBr 3 The Al to Br 2 molar ratio of 2:3 will be used to answer (a). The Al to AlBr 3 molar ratio of 2:2 will be used to answer (b).

2   0.0054037 mol ––––  =  ––––––––––––– 3   x x = 0.00810555 mol (of Br 2 )

(0.00810555 mol) (159.808 g/mol) = 1.30 g (to three sig figs)

2   0.0054037 mol ––––  =  ––––––––––––– 2   x x = 0.0054037 mol (of AlBr 3 )

(0.0054037 mol) (266.694 g/mol) = 1.44 g (to three sig figs)

66.0 g / 40.078 g/mol = 1.6468 mol

3   8 –––––––––  =  ––– 1.6368 mol   x x = 4.3648 mol

(4.3648 mol) (16.00 g/mol) = 69.8 g

2LiOH + CO 2 ---> Li 2 CO 3 + H 2 O

2Li + 2H 2 O ---> 2LiOH + H 2

2Li + CO 2 + H 2 O ---> Li 2 CO 3 + H 2 Note that two LiOH and one H 2 O cancel out. This third reaction gives me the Li to CO 2 as 2 to 1, so I am now ready to continue on.

1000 g / 6.941 g/mol = 144.07 mol

2   144.07 mol –––  =  ––––––––– 1   x x = 72.035 mol (of CO 2 )

(72.035 mol) (44.009 g/mol) = 3170 g

PV = nRT (1.00 atm) (V) = (72.035 mol) (0.08206 L atm / mol K) (273.15 K) V = 1614.6 L (to three sig figs, this would be 1610 L)

(22.414 L/mol) (72.035 mol) = 1614.6 L (1610 L to three sig figs)