Chapter 16 — Front Matter
Purpose and Overview
Every recipe uses ratios. If you want to make two dozen cookies instead of one dozen, you double every ingredient. If you want to make only half a batch, you cut every ingredient in half. The relationships between ingredients are fixed by the recipe itself. Chemical reactions work the same way.
A balanced chemical equation is a recipe. It tells you exactly how much of each substance reacts and exactly how much of each product forms. The ratios between substances are fixed by the balanced equation. Stoichiometry is the branch of chemistry that reads those ratios and uses them to make quantitative predictions about chemical reactions.
The word stoichiometry comes from Greek words meaning "element" and "measure." Stoichiometry is, quite literally, the measurement of chemical substances in reactions. In this chapter, you will learn to predict how much product a reaction can form, determine which reactant runs out first, and measure how efficient a real reaction is compared to what theory predicts.
Opening Question
A pharmaceutical company synthesizes aspirin by reacting salicylic acid with acetic anhydride. Before running a batch, the chemists must calculate exactly how much of each starting material is needed, which reactant will run out first, and how much product they can expect. If they calculate wrong, they waste expensive materials or fall short of their target. By the end of this chapter, you will have every tool needed to answer these questions for any reaction.
10Sections
9Objectives
7Key Terms
13+Worked Examples
Learning Objectives
- 1Explain what stoichiometry studies and why it is essential in chemistry.
- 2Read mole ratios from balanced chemical equations.
- 3Use mole ratios as conversion factors between different substances in a reaction.
- 4Perform mole-to-mole stoichiometric calculations.
- 5Perform mole-to-mass stoichiometric calculations.
- 6Perform mass-to-mass stoichiometric calculations.
- 7Identify the limiting reactant and calculate the theoretical yield of product.
- 8Calculate how much excess reactant remains after a reaction.
- 9Calculate percent yield and explain why actual yield is less than theoretical yield.
Sections 16.1 – 16.3
Foundations: What Stoichiometry Is and How Mole Ratios Work
Purpose · Equations as Ratios · Mole Ratios as Conversion Factors
Section 16.1
What Stoichiometry Studies
Chemists do not only ask what happens in a reaction. They ask how much. How many grams of iron are needed to produce a specific mass of iron oxide? How much carbon dioxide forms when a fuel burns? How much aspirin can be synthesized from a given quantity of salicylic acid? These questions are answered by stoichiometry.
Stoichiometry rests on one central idea: balanced chemical equations represent fixed ratios between particles, and those ratios allow chemists to calculate amounts of substances involved in reactions. When you balance a chemical equation, you are not just satisfying conservation of mass. You are establishing a set of mole ratios — the proportions in which reactants combine and products form.
The Four Calculation Types
| Calculation Type | Given | Find | Conversions Needed |
|---|
| Mole-to-mole | mol of substance A | mol of substance B | Mole ratio only |
| Mole-to-mass | mol of substance A | g of substance B | Mole ratio, then molar mass of B |
| Mass-to-mass | g of substance A | g of substance B | Molar mass of A, mole ratio, molar mass of B |
| Limiting reactant | amounts of two or more reactants | theoretical yield of product | All of the above, applied to each reactant |
Section 16.2
Balanced Equations as Ratios
Consider the synthesis of water:
2 H₂ + O₂ → 2 H₂O
The coefficients 2, 1, and 2 can be read at the particle level or the mole level. At the particle level: 2 molecules of hydrogen react with 1 molecule of oxygen to produce 2 molecules of water. At the mole level: 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. Both readings are correct. The ratios are identical whether you are counting particles or counting in moles.
Reading Mole Ratios from Equations
Consider the synthesis of ammonia:
N₂ + 3 H₂ → 2 NH₃
| Substances | Mole Ratio (as fraction) |
|---|
| N₂ to H₂ | 1 mol N₂ / 3 mol H₂ |
| H₂ to N₂ | 3 mol H₂ / 1 mol N₂ |
| N₂ to NH₃ | 1 mol N₂ / 2 mol NH₃ |
| NH₃ to N₂ | 2 mol NH₃ / 1 mol N₂ |
| H₂ to NH₃ | 3 mol H₂ / 2 mol NH₃ |
| NH₃ to H₂ | 2 mol NH₃ / 3 mol H₂ |
Choosing the Correct Mole Ratio
Always place the unknown substance in the numerator and the given substance in the denominator. This ensures the given units cancel and the unknown units remain after multiplication.
Example 16.2.1
Using a mole ratio — N₂ + 3 H₂ → 2 NH₃
- ProblemHow many moles of NH₃ form from 6 moles of H₂?
- Given6 mol H₂ · Unknown: mol NH₃
- Ratio
2 mol NH₃ / 3 mol H₂ (unknown on top, given on bottom)
- Solve
6 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 4 mol NH₃
Answer: 4 mol NH₃
Section 16.3
Mole Ratios
The mole ratio serves as a conversion factor between moles of different substances in a reaction — exactly as molar mass converts between moles and grams. When you multiply by a mole ratio, you are converting from one substance to another while preserving the proportional relationship dictated by the equation.
General Setup
mol given × (mol unknown / mol given) = mol unknown
Always write units explicitly and confirm they cancel before doing arithmetic.
Example 16.3.1
Moles of O₂ needed — CH₄ + 2 O₂ → CO₂ + 2 H₂O
- ProblemHow many moles of O₂ are needed to react completely with 5.0 mol of CH₄?
- Ratio
2 mol O₂ / 1 mol CH₄
- Solve
5.0 mol CH₄ × (2 mol O₂ / 1 mol CH₄) = 10.0 mol O₂
Answer: 10.0 mol O₂
Example 16.3.2
Moles of H₂O produced — same equation
- ProblemHow many moles of H₂O form from 5.0 mol of CH₄?
- Ratio
2 mol H₂O / 1 mol CH₄
- Solve
5.0 mol CH₄ × (2 mol H₂O / 1 mol CH₄) = 10.0 mol H₂O
Answer: 10.0 mol H₂O
Example 16.3.3
Moles of CH₄ needed to produce a given amount of CO₂
- ProblemHow many moles of CH₄ are needed to produce 8.4 mol of CO₂?
- Ratio
1 mol CH₄ / 1 mol CO₂
- Solve
8.4 mol CO₂ × (1 mol CH₄ / 1 mol CO₂) = 8.4 mol CH₄
- CheckThe 1:1 ratio between CH₄ and CO₂ in the equation confirms the answer should equal the given moles. ✓
Answer: 8.4 mol CH₄
Reasonableness Check
After every mole ratio calculation, ask whether the answer makes sense with the coefficients. If the equation shows a 2:1 ratio and you expect more product than reactant, confirm your number is larger, not smaller. Catching direction errors this way saves time on every problem.
Sections 16.4 – 16.6
The Three Stoichiometric Calculation Types
Mole-to-Mole · Mole-to-Mass · Mass-to-Mass
Section 16.4
Mole-to-Mole Stoichiometry
Mole-to-mole stoichiometry is the most direct calculation type. You are given moles of one substance and asked to find moles of another. The only conversion factor needed is the mole ratio from the balanced equation. This is the conceptual core of all stoichiometry — every more complex calculation contains a mole-to-mole step at its center.
Mole-to-Mole Pathway
mol given
× mole ratio →
mol unknown
Use the equation 4 Fe + 3 O₂ → 2 Fe₂O₃ for all problems.
- How many moles of O₂ are needed to react with 12.0 mol of Fe?
- How many moles of Fe₂O₃ are produced from 6.0 mol of Fe?
- How many moles of Fe are needed to produce 5.0 mol of Fe₂O₃?
- How many moles of O₂ are needed to produce 10.0 mol of Fe₂O₃?
Answers: 1. 9.0 mol O₂ · 2. 3.0 mol Fe₂O₃ · 3. 10.0 mol Fe · 4. 15.0 mol O₂
Section 16.5
Mole-to-Mass Stoichiometry
When you are given moles of one substance and asked for the mass in grams of another, you add one step to the pathway: convert moles of the unknown to grams using its molar mass.
Mole-to-Mass Pathway
mol given
× mole ratio →
mol unknown
× molar mass →
g unknown
Example 16.5.1
Grams of Fe₂O₃ from moles of Fe — 4 Fe + 3 O₂ → 2 Fe₂O₃
- ProblemHow many grams of Fe₂O₃ are produced from 2.0 mol of Fe?
- Molar MassFe₂O₃: 2(55.85) + 3(16.00) = 111.70 + 48.00 = 159.70 g/mol
- Setup
2.0 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe) × (159.70 g Fe₂O₃ / 1 mol Fe₂O₃)
- Calculate
2.0 × 0.500 × 159.70 = 159.7 g → 160 g (2 sig figs)
Answer: 160 g Fe₂O₃
Example 16.5.2
Grams of O₂ needed to burn propane — C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
- ProblemHow many grams of O₂ are required to burn 3.0 mol of C₃H₈?
- Molar MassO₂: 2(16.00) = 32.00 g/mol
- Setup
3.0 mol C₃H₈ × (5 mol O₂ / 1 mol C₃H₈) × (32.00 g O₂ / 1 mol O₂)
- Calculate
3.0 × 5 × 32.00 = 480 g O₂
Answer: 480 g O₂
Significant Figures Reminder
Carry at least four significant figures in molar mass values from the periodic table. Apply significant figures to the final answer only, not to intermediate steps. Rounding too early accumulates error.
Section 16.6
Mass-to-Mass Stoichiometry
In real laboratory work, quantities are almost always measured in grams. Mass-to-mass stoichiometry is therefore the most commonly used form. The key insight is this: you cannot directly convert grams of one substance to grams of another. You must always pass through moles. Moles are the bridge between substances in a chemical equation.
Mass-to-Mass Pathway (Three Steps)
g given
÷ molar mass A →
mol given
× mole ratio →
mol unknown
× molar mass B →
g unknown
Example 16.6.1
Grams of H₂O from grams of glucose — C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
- ProblemHow many grams of water are produced when 36.0 g of glucose undergoes cellular respiration?
- Molar MassesC₆H₁₂O₆ = 180.16 g/mol · H₂O = 18.02 g/mol
- Setup
36.0 g C₆H₁₂O₆ × (1 mol / 180.16 g) × (6 mol H₂O / 1 mol C₆H₁₂O₆) × (18.02 g / 1 mol H₂O)
- Calculate
= 0.1998 × 6 × 18.02 = 21.6 g H₂O
- Check36.0 g of glucose is 0.200 mol. Six moles of water form per mole of glucose, so 1.20 mol H₂O × 18.02 g/mol = 21.6 g. ✓
Answer: 21.6 g H₂O
Example 16.6.2
Grams of AlCl₃ from grams of Al — 2 Al + 3 Cl₂ → 2 AlCl₃
- ProblemHow many grams of AlCl₃ are produced when 54.0 g of aluminum reacts with excess Cl₂?
- Molar MassesAl = 26.98 g/mol · AlCl₃ = 26.98 + 3(35.45) = 133.33 g/mol
- Setup
54.0 g Al × (1 mol Al / 26.98 g) × (2 mol AlCl₃ / 2 mol Al) × (133.33 g AlCl₃ / 1 mol AlCl₃)
- Calculate
= 2.001 mol Al × 1 × 133.33 = 266.9 g → 267 g (3 sig figs)
Answer: 267 g AlCl₃
Never Skip the Mole Bridge
Even when the mole ratio is 1:1, you must still pass through moles. Multiplying grams of one substance by a mass ratio will give wrong answers unless the molar masses of the two substances are identical. The mole ratio is the only valid bridge between substances in a reaction.
Sections 16.7 – 16.8
Limiting and Excess Reactants
Identifying the Limiting Reactant · Calculating Excess Remaining · Mass-Based Problems
Section 16.7
Limiting Reactants
So far, you have assumed that all reactants are available in exactly the right proportions for complete reaction. In real situations, this is rarely the case. Imagine you are assembling sandwiches. Each sandwich requires 2 slices of bread and 1 slice of cheese. If you have 10 slices of bread and 4 slices of cheese, you can make only 4 sandwiches — the cheese runs out first, regardless of how much bread remains. Chemical reactions work exactly the same way.
Definitions
Limiting reactant — the reactant that is completely consumed first. It determines the maximum amount of product that can form. When it is used up, the reaction stops.
Excess reactant — any reactant present in greater amount than needed. Some remains unreacted after the limiting reactant is consumed.
Method: Convert Each Reactant to Moles of Product
For each reactant, calculate how many moles of product it could produce if it were the only limiting factor. The reactant that produces the least product is the limiting reactant. This method is systematic and works for all situations.
Example 16.7.1
Identifying the limiting reactant — N₂ + 3 H₂ → 2 NH₃
- ProblemYou have 4.0 mol N₂ and 9.0 mol H₂. Which is the limiting reactant? How many moles of NH₃ can form?
- From N₂
4.0 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 8.0 mol NH₃
- From H₂
9.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 6.0 mol NH₃
- CompareH₂ produces fewer moles of NH₃ (6.0 < 8.0). H₂ is the limiting reactant.
Limiting reactant: H₂ · Product formed: 6.0 mol NH₃
Example 16.7.2
Identifying the limiting reactant — 2 H₂ + O₂ → 2 H₂O
- ProblemYou have 6.0 mol H₂ and 4.0 mol O₂. Which is limiting?
- From H₂
6.0 mol H₂ × (2 mol H₂O / 2 mol H₂) = 6.0 mol H₂O
- From O₂
4.0 mol O₂ × (2 mol H₂O / 1 mol O₂) = 8.0 mol H₂O
- CompareH₂ produces fewer moles of H₂O (6.0 < 8.0). H₂ is the limiting reactant.
Limiting reactant: H₂ · Product formed: 6.0 mol H₂O
Common Misconception
The reactant present in the smaller number of moles is not always the limiting reactant. The limiting reactant depends on both the amount present and the mole ratio required by the equation. Always calculate moles of product from each reactant — never simply compare mole quantities.
Section 16.8
Excess Reactants
After the limiting reactant is consumed, some of the excess reactant remains unreacted. Chemists need to know how much in order to recover and reuse it, or to account for it in safety and waste calculations.
Calculating Excess Remaining
mol excess remaining = mol excess starting − mol excess consumed
First identify the limiting reactant. Then calculate how much excess reactant was consumed by the limiting reactant. Subtract from the starting amount.
Example 16.8.1
Excess N₂ remaining — N₂ + 3 H₂ → 2 NH₃
- SetupFrom Example 16.7.1: 4.0 mol N₂ and 9.0 mol H₂. H₂ is the limiting reactant.
- N₂ Consumed
9.0 mol H₂ × (1 mol N₂ / 3 mol H₂) = 3.0 mol N₂ consumed
- N₂ Remaining
4.0 mol N₂ − 3.0 mol N₂ = 1.0 mol N₂ remaining
1.0 mol N₂ remains unreacted after the reaction.
Example 16.8.2
Mass-based limiting reactant — 2 Al + 3 Cl₂ → 2 AlCl₃
- ProblemYou have 27.0 g of Al and 106.5 g of Cl₂. Which is the limiting reactant? How many grams of AlCl₃ form?
- Convert
mol Al = 27.0 / 26.98 = 1.00 mol
mol Cl₂ = 106.5 / 70.90 = 1.50 mol
- From Al
1.00 mol Al × (2 mol AlCl₃ / 2 mol Al) = 1.00 mol AlCl₃
- From Cl₂
1.50 mol Cl₂ × (2 mol AlCl₃ / 3 mol Cl₂) = 1.00 mol AlCl₃
- ResultBoth produce equal product — the stoichiometric ratio is exact. Neither is in excess. Product = 1.00 mol AlCl₃ × 133.33 g/mol = 133 g AlCl₃.
Answer: 133 g AlCl₃ · No excess reactant (exact stoichiometric ratio)
Sections 16.9 – 16.10
Yield and Applications
Theoretical Yield · Actual Yield · Percent Yield · Industry and Laboratories
Section 16.9
Theoretical Yield, Actual Yield, and Percent Yield
Stoichiometric calculations give you the theoretical yield — the maximum amount of product that could form if the reaction were perfectly complete, all substances were pure, no product was lost in handling, and no side reactions occurred. In real laboratory and industrial situations, the amount of product actually collected — the actual yield — is almost always less.
Why Actual Yield Falls Short
| Reason | Explanation |
|---|
| Incomplete reaction | The reaction does not go to 100% conversion |
| Product losses | Material lost during transfer between containers |
| Side reactions | Reactants consumed by competing reactions that form different products |
| Equilibrium | Some reactions reach a balance point before completion |
| Purification losses | Some product lost during filtering, crystallization, or drying |
Example 16.9.1
Percent yield — aspirin synthesis
- ProblemStoichiometry predicts 14.4 g of aspirin. A student collects 11.7 g. What is the percent yield?
- Solve
Percent Yield = (11.7 g / 14.4 g) × 100% = 81.3%
Answer: 81.3% — the student collected about 81% of the maximum possible product.
Example 16.9.2
Theoretical yield then percent yield — combustion of ethanol
- Equation
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
- ProblemA chemist burns 23.0 g of ethanol and collects 40.5 g of CO₂. Find the theoretical yield and the percent yield.
- Molar MassesC₂H₅OH = 46.07 g/mol · CO₂ = 44.01 g/mol
- Theoretical Yield
23.0 g C₂H₅OH × (1 mol / 46.07 g) × (2 mol CO₂ / 1 mol C₂H₅OH) × (44.01 g / 1 mol CO₂)
= 0.4993 × 2 × 44.01 = 43.9 g CO₂
- Percent Yield
(40.5 / 43.9) × 100% = 92.2%
Theoretical yield: 43.9 g CO₂ · Percent yield: 92.2%
Industrial Significance
In pharmaceutical manufacturing, a five-step synthesis where each step has 90% yield produces a final overall yield of approximately 59% (0.90⁵ ≈ 0.59). Improving even one step significantly increases total production and reduces material waste and cost.
Section 16.10
Stoichiometry in Industry and Laboratories
Every major chemical industry depends on stoichiometric calculations. Without them, manufacturers would have no way to know how much raw material to purchase, how much waste to expect, or how efficiently their processes are running.
| Application | Stoichiometric Role |
|---|
| Haber-Bosch Process (ammonia for fertilizers) | Maintains N₂ : H₂ mole ratios; controls limiting reactant; measures percent yield continuously |
| Pharmaceutical synthesis | Calculates batch sizes; multiplies step yields across multi-step syntheses |
| Combustion engineering | Optimizes fuel-to-air ratio for complete combustion; minimizes CO and unburned fuel |
| Water treatment | Calculates mass of disinfectant needed per volume of water |
| Environmental monitoring | Calculates CO₂ produced per liter of fuel burned; models atmospheric chemistry |
| Laboratory planning | Determines how much reactant is needed to produce a target amount of product |
Chemistry Connection
The Haber-Bosch process — N₂ + 3 H₂ → 2 NH₃ — is one of the most important chemical reactions in human history. It produces the ammonia used in fertilizers that support the agricultural production feeding billions of people. The precise stoichiometric control of this reaction, developed in the early twentieth century, transformed global food production.
Chapter 16 — Back Matter
Chapter Summary
- Sec 16.1Stoichiometry is the quantitative study of chemical reactions. Balanced equations encode fixed mole ratios that allow chemists to calculate amounts of any substance from any other substance in the reaction.
- Sec 16.2Coefficients in balanced equations represent particle ratios and mole ratios simultaneously. Every pair of substances in an equation produces a mole ratio usable as a conversion factor. The unknown goes in the numerator; the given goes in the denominator.
- Sec 16.3Mole ratios convert between moles of different substances. Always write units explicitly and confirm they cancel. Check answers for reasonableness against the equation's coefficients.
- Sec 16.4Mole-to-mole stoichiometry uses the mole ratio directly. It is the conceptual core of all stoichiometric calculations.
- Sec 16.5Mole-to-mass stoichiometry adds conversion of mol unknown to g unknown using molar mass. The pathway is: mol given → mol unknown → g unknown.
- Sec 16.6Mass-to-mass stoichiometry is the most common type. The pathway is: g given → mol given → mol unknown → g unknown. Moles are the required bridge; never convert grams directly to grams.
- Sec 16.7The limiting reactant is completely consumed first and sets the maximum product yield. To identify it, calculate moles of product each reactant could produce and choose the one giving the least. Do not simply compare mole quantities.
- Sec 16.8Excess reactant remaining = moles starting − moles consumed. Use the limiting reactant to calculate how much of the excess reactant was consumed, then subtract. Mass-based problems require converting grams to moles first.
- Sec 16.9Theoretical yield is the maximum product predicted by stoichiometry. Actual yield is what is collected. Percent yield = (actual / theoretical) × 100%. Values above 100% are impossible; they indicate measurement error or contamination.
- Sec 16.10Stoichiometry is used in every branch of chemistry and chemical industry — from the Haber-Bosch process and pharmaceutical manufacturing to combustion engineering, water treatment, and environmental monitoring.
Key Terms
stoichiometry
The branch of chemistry that uses balanced equations to calculate quantitative relationships between reactants and products in a chemical reaction.
mole ratio
A ratio derived from the coefficients of a balanced chemical equation; used as a conversion factor between moles of different substances.
limiting reactant
The reactant that is completely consumed first and determines the maximum amount of product that can form. Also called the limiting reagent.
excess reactant
A reactant present in greater quantity than the stoichiometric ratio requires; some of it remains unreacted after the limiting reactant is consumed.
theoretical yield
The maximum amount of product calculated from stoichiometry, assuming complete reaction with no losses and no side reactions.
actual yield
The amount of product actually collected from a reaction in practice; always less than or equal to the theoretical yield.
percent yield
The ratio of actual yield to theoretical yield, multiplied by 100%; a measure of reaction efficiency. Values above 100% are impossible.
Formula Review
Worked Example Review
Worked Example A — Mole-to-Mole
2 SO₂ + O₂ → 2 SO₃ · How many moles of SO₃ form from 5.0 mol of SO₂?
- Solve
5.0 mol SO₂ × (2 mol SO₃ / 2 mol SO₂) = 5.0 mol SO₃
Answer: 5.0 mol SO₃
Worked Example B — Mass-to-Mass
2 H₂ + O₂ → 2 H₂O · How many grams of water form from 8.0 g of H₂?
- Molar MassesH₂ = 2.016 g/mol · H₂O = 18.02 g/mol
- Solve
8.0 g H₂ × (1 mol / 2.016 g) × (2 mol H₂O / 2 mol H₂) × (18.02 g / 1 mol H₂O)
= 3.968 × 1 × 18.02 = 71.5 g H₂O
Answer: 71.5 g H₂O
Worked Example C — Limiting Reactant
Fe + S → FeS · Given 3.0 mol Fe and 4.0 mol S
- From Fe
3.0 mol Fe × (1 mol FeS / 1 mol Fe) = 3.0 mol FeS
- From S
4.0 mol S × (1 mol FeS / 1 mol S) = 4.0 mol FeS
- LimitingFe produces less product. Fe is the limiting reactant.
- Excess S
4.0 mol − 3.0 mol = 1.0 mol S remaining
Limiting: Fe · Product: 3.0 mol FeS · Excess: 1.0 mol S
Worked Example D — Percent Yield
Theoretical yield = 25.0 g · Actual yield = 19.5 g
- Solve
(19.5 / 25.0) × 100% = 78.0%
Answer: 78.0% yield
Concept Check
Answer these questions in writing before attempting the full practice problems.
- What is the difference between a subscript and a coefficient in a chemical equation? Which one changes when you balance an equation?
- Explain in your own words what a mole ratio is and where it comes from.
- A balanced equation shows 3 mol of reactant A producing 2 mol of product B. If you have 6 mol of A, how many mol of B form? Show your reasoning before calculating.
- Why must you always convert grams to moles before applying a mole ratio?
- In your own words, explain why the limiting reactant is called "limiting."
- A reaction has a theoretical yield of 40.0 g but only 30.0 g of product was collected. Is the percent yield above or below 100%? Calculate it.
- Why is a percent yield greater than 100% impossible?
Stoichiometry Practice
Use the equation 2 KClO₃ → 2 KCl + 3 O₂ for problems 1–4.
- Write the mole ratio of KClO₃ to O₂.
- Write the mole ratio of O₂ to KCl.
- How many moles of O₂ form from 4.0 mol of KClO₃?
- How many moles of KClO₃ are needed to produce 9.0 mol of O₂?
Use the equation 2 Na + Cl₂ → 2 NaCl for problems 5–8.
- Moles of NaCl from 6.0 mol Na?
- Moles of Cl₂ needed for 10.0 mol Na?
- Moles of Na needed for 3.5 mol NaCl?
- Moles of NaCl from 2.5 mol Cl₂?
- S + O₂ → SO₂ — How many grams of SO₂ form from 32.0 g of sulfur?
- 2 Mg + O₂ → 2 MgO — How many grams of MgO form from 12.2 g of Mg?
- N₂ + 3 H₂ → 2 NH₃ — How many grams of NH₃ form from 42.0 g of N₂?
- CH₄ + 2 O₂ → CO₂ + 2 H₂O — How many grams of CO₂ form from 80.0 g of CH₄?
- 2 Al + 3 Br₂ → 2 AlBr₃ — How many grams of Br₂ are needed to react completely with 13.5 g of Al?
- 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O — Starting with 68.0 g of NH₃, calculate the grams of H₂O produced.
- CaCO₃ → CaO + CO₂ — If 250.0 g of CaCO₃ decomposes, what is the theoretical yield of CaO in grams?
Limiting Reactant Practice
For each problem: identify the limiting reactant, calculate the theoretical yield of the specified product, and calculate how much excess reactant remains.
- Starting amounts: 4.0 mol H₂ and 3.0 mol O₂. Find: mol H₂O formed and mol excess reactant remaining.
- Starting amounts: 6.0 mol H₂ and 2.0 mol O₂. Find: mol H₂O formed and mol excess reactant remaining.
- Starting amounts: 5.0 mol N₂ and 12.0 mol H₂. Which is limiting? How many moles of NH₃ form?
- Starting amounts: 28.0 g N₂ and 9.0 g H₂. Which is limiting? How many grams of NH₃ form?
- Starting amounts: 160.0 g Fe₂O₃ and 100.0 g CO. Which is limiting? What is the theoretical yield of Fe in grams?
Percent Yield Practice
- A student prepares copper(II) sulfate. Theoretical yield: 19.9 g. Actual yield: 17.1 g. Calculate the percent yield.
- A reaction has a theoretical yield of 56.0 g of product. If the percent yield is 74.0%, what is the actual yield?
- A chemist produces 45.2 g of a compound. If the percent yield was 88.0%, what was the theoretical yield?
- A synthesis of a dye molecule gives 32.5 g of product. Stoichiometric calculations predict 40.0 g. Calculate the percent yield and suggest two possible reasons why it is less than 100%.
- A pharmaceutical reaction runs at 91.5% yield. If the theoretical yield is 500.0 g, how many grams of product are collected?
Error Analysis
Each problem contains a stoichiometric error. Identify the error and correct it.
Equation for Problems A and B: 2 HCl + Ca → CaCl₂ + H₂
Problem A
4.0 mol HCl × (2 mol H₂ / 1 mol HCl) = 8.0 mol H₂
The mole ratio is inverted and uses incorrect coefficients. What should it be?
Problem B
6.0 mol HCl × (110.98 g CaCl₂ / 1 mol HCl) = 665.9 g CaCl₂
The mole ratio step was skipped entirely. What is the correct calculation?
Problem C
Percent yield = (24.0 / 18.0) × 100% = 133%
The formula was set up upside down. Correct it and explain why a value above 100% is impossible.
Problem D
"2 Na + Cl₂ → 2 NaCl. Given 3.0 mol Na and 2.0 mol Cl₂. There are fewer moles of Na, so Na is the limiting reactant."
The student compared mole quantities instead of calculating moles of product. Show the correct method.
Stoichiometry in Industry and Laboratories
- A plant produces ammonia by the Haber process: N₂ + 3 H₂ → 2 NH₃. If 1000 kg of N₂ is fed into the reactor, what is the theoretical yield of NH₃ in kilograms? (Convert kg to g first, then to mol, then apply stoichiometry.)
- A pharmaceutical company synthesizes a drug through a 4-step process with percent yields of 95%, 88%, 79%, and 91%. Calculate the overall percent yield by multiplying the decimal form of each step's yield.
- An automobile burns octane: 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O. How many grams of CO₂ are emitted when 500.0 g of octane burns completely?
- A water treatment facility adds Cl₂ gas at 7.1 kg per day. If the disinfection reaction operates at 95.0% yield with a 1:1 mole ratio between Cl₂ and the active product, how many kg of active disinfectant product forms daily?
Mixed Review
Select and combine stoichiometric skills without being told which method to use.
- 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O — Starting with 58.2 g of butane (C₄H₁₀), how many grams of CO₂ form?
- Zn + 2 HCl → ZnCl₂ + H₂ — Starting with 32.5 g of Zn and 20.0 g of HCl, identify the limiting reactant and calculate the theoretical yield of ZnCl₂ in grams.
- A student runs the reaction in problem 2 and collects 52.8 g of ZnCl₂. Calculate the percent yield.
- 4 P + 5 O₂ → 2 P₂O₅ — Starting with 6.2 mol P and 8.0 mol O₂: (a) Which is limiting? (b) How many moles of P₂O₅ form? (c) How much excess reactant remains in moles?
- A reaction produces 88.5 g of product with a percent yield of 73.2%. What was the theoretical yield?
Chapter 16 Checkpoint
Chapter 16 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 16
Section 1 — Vocabulary and Concepts
- Define stoichiometry in your own words and explain why it requires a balanced equation.
- What is a mole ratio and where does it come from? Give an example using:
2 H₂ + O₂ → 2 H₂O
- Explain the difference between the limiting reactant and the excess reactant.
- Why is actual yield almost always less than theoretical yield? Give two reasons.
Section 2 — Mole Ratio and Mole-to-Mole
- Using
N₂ + 3 H₂ → 2 NH₃, write all six possible mole ratios between any two substances.
- How many moles of NH₃ form from 9.0 mol H₂?
- How many moles of N₂ are needed to produce 10.0 mol NH₃?
Section 3 — Mass-to-Mass Calculations
- Using
2 Al + 3 Cl₂ → 2 AlCl₃, how many grams of AlCl₃ form from 81.0 g of Al?
- Using
C + O₂ → CO₂, how many grams of CO₂ form from 24.0 g of carbon?
Section 4 — Limiting Reactants
- Given
2 H₂ + O₂ → 2 H₂O with 5.0 mol H₂ and 3.0 mol O₂: identify the limiting reactant, state the theoretical yield in moles, and calculate the excess reactant remaining.
- Given
Fe + S → FeS with 28.0 g Fe and 20.0 g S: which is the limiting reactant and how many grams of FeS form?
Section 5 — Percent Yield
- A student synthesizes MgO and collects 14.3 g. The theoretical yield was 16.1 g. Calculate the percent yield.
- A reaction has a theoretical yield of 80.0 g and runs at 87.5% yield. How many grams of product are collected?
Section 6 — Error Analysis
- A student calculates percent yield as (theoretical / actual) × 100% and gets 122%. What went wrong? What is the likely correct answer?
- A student concludes that 2.0 mol of reactant A is the limiting reactant because it is less than the 5.0 mol of reactant B present, without doing any calculation. Explain why this reasoning is flawed and describe the correct method.
Looking Ahead
Chapter 17: Reactions in Aqueous Solution
Chapter 17 moves stoichiometry into the aqueous world. Most chemical reactions in biology, medicine, geology, and environmental science take place in water. When ionic compounds dissolve, they separate into individual ions that can react with other ions. You will learn about dissociation, electrolytes, precipitation reactions, net ionic equations, and concentration expressed as molarity — and you will apply the stoichiometric skills from this chapter to reactions in solution. Understanding how ions behave in water prepares you for acid-base chemistry, titration, and environmental chemistry in later chapters.
MGU Library Connection
- Chapter 15 — The Mole and Molar Mass: mole conversions required for stoichiometry
- Chapter 13 — Chemical Equations: balancing equations that produce the mole ratios used here
- Chapter 17 — Reactions in Aqueous Solution: stoichiometry applied to dissolved substances and titration
- Appendix H — Gas Law and Thermochemistry Formula Sheet: companion formulas for quantitative chemistry
- Appendix K — Selected Answers: odd-numbered problems from this chapter