Stoichiometry Practice Problems: Moles, Mass, Ratios, and Limiting Reactants

Overview

Stoichiometry is the quantitative study of relationships between substances in a chemical reaction. A balanced equation provides the mole ratio, while molar mass connects mass and moles. Most chemistry practice problems are variations of one central workflow:

1. Write and balance the chemical equation.
2. Record the given quantity and the requested quantity.
3. Convert the given measurement to moles if necessary.
4. Use the coefficients in the balanced equation as a mole ratio.
5. Convert the answer into the unit requested by the problem.
6. Check units, significant figures, and whether the answer is chemically reasonable.

The most useful general setup is dimensional analysis. For example, a mass-to-mass calculation can be organized as:

grams of A × (1 mol A / molar mass of A) × (mol B / mol A) × (molar mass of B / 1 mol B) = grams of B

The units should cancel step by step, leaving only the desired unit. If they do not, stop and correct the setup before calculating.

Core formulas and ideas

• Moles from mass: moles = mass in grams ÷ molar mass in grams per mole.
• Mass from moles: mass in grams = moles × molar mass.
• Mole ratio: use coefficients from the balanced equation, not subscripts in chemical formulas.
• Percent yield: percent yield = (actual yield ÷ theoretical yield) × 100.
• Limiting reactant: the reactant that produces the smaller amount of product and is consumed first.

Checklist by scenario

Scenario 1: Mole-to-mole conversions

Suppose nitrogen reacts with hydrogen according to:

N₂ + 3H₂ → 2NH₃

How many moles of ammonia can form from 4.50 mol H₂?

1. Confirm the equation is balanced.
2. Start with the given amount: 4.50 mol H₂.
3. Apply the coefficient ratio: 2 mol NH₃ for every 3 mol H₂.
4. Calculate: 4.50 mol H₂ × (2 mol NH₃ ÷ 3 mol H₂) = 3.00 mol NH₃.

Because both quantities are already in moles, no molar-mass conversion is needed.

Scenario 2: Mass-to-mole conversions

How many moles are present in 18.0 g of water, H₂O? Using a molar mass of approximately 18.0 g/mol:

18.0 g H₂O × (1 mol H₂O ÷ 18.0 g H₂O) = 1.00 mol H₂O

For this type of problem, the reaction equation is not necessary unless the question asks you to relate the water to another substance.

Scenario 3: Mass-to-mass stoichiometry

Consider the decomposition of calcium carbonate:

CaCO₃ → CaO + CO₂

How many grams of CO₂ form from 50.0 g CaCO₃?

1. Convert calcium carbonate to moles. Its molar mass is approximately 100.1 g/mol.
2. Use the 1:1 mole ratio between CaCO₃ and CO₂.
3. Convert moles of CO₂ to grams using its molar mass, approximately 44.0 g/mol.

50.0 g CaCO₃ × (1 mol CaCO₃ ÷ 100.1 g CaCO₃) × (1 mol CO₂ ÷ 1 mol CaCO₃) × (44.0 g CO₂ ÷ 1 mol CO₂) = 22.0 g CO₂

Do not skip the middle mole step, even when the equation has a simple ratio. It keeps the reasoning visible and helps prevent unit errors.

Scenario 4: Percent yield

In a laboratory, the calculated theoretical yield is 12.5 g, but the experiment produces 10.0 g. The percent yield is:

(10.0 g ÷ 12.5 g) × 100 = 80.0%

The theoretical yield comes from stoichiometry. The actual yield comes from the measured experiment. Keep these values separate and use the same units before dividing.

Scenario 5: Limiting reactant problems

For a reaction with more than one reactant, do not assume that the reactant with the smaller mass is limiting. Convert each reactant to moles and calculate how much product each could produce.

For example:

2H₂ + O₂ → 2H₂O

Assume you have 5.00 mol H₂ and 2.00 mol O₂.

• From hydrogen: 5.00 mol H₂ × (2 mol H₂O ÷ 2 mol H₂) = 5.00 mol H₂O.
• From oxygen: 2.00 mol O₂ × (2 mol H₂O ÷ 1 mol O₂) = 4.00 mol H₂O.

Oxygen produces less water, so O₂ is the limiting reactant. The theoretical yield is therefore 4.00 mol H₂O, not 5.00 mol.

Practice set

1. How many moles of O₂ are needed to react completely with 6.00 mol H₂ in the equation 2H₂ + O₂ → 2H₂O? Answer: 3.00 mol O₂.
2. How many grams of MgO form from 12.0 g Mg in 2Mg + O₂ → 2MgO? Answer: approximately 19.9 g MgO.
3. If the theoretical yield is 8.40 g and the actual yield is 7.14 g, what is the percent yield? Answer: 85.0%.
4. In N₂ + 3H₂ → 2NH₃, which reactant limits production when 2.00 mol N₂ reacts with 4.00 mol H₂? Answer: H₂, because 2.00 mol N₂ requires 6.00 mol H₂.

What to double-check

• Equation balance: Count each element on both sides. Never alter subscripts to balance an equation; change coefficients instead.
• Molar mass: Add the atomic masses for every atom in the formula, including repeated atoms indicated by parentheses.
• Coefficient ratio: Coefficients describe mole relationships. Subscripts describe the composition of individual particles.
• Unit cancellation: Write units beside every number. A correct numerical answer with incorrect units is not a complete solution.
• Limiting reactant logic: Compare possible product amounts or compare the amount required with the amount available. The smaller product amount identifies the limiting reactant.
• Significant figures: Report the final result to a precision supported by the given measurements and the molar masses used in class.
• Excess reactant: Once the limiting reactant is identified, only it determines the maximum product. The other reactant remains in excess, although its leftover amount can also be calculated.

Common mistakes

Using an unbalanced equation: A coefficient ratio is meaningful only after the equation is balanced. Balance first, calculate second.

Using grams directly in the mole ratio: Chemical equations compare moles, not grams. Convert grams to moles before applying coefficients.

Reversing the ratio: If the equation shows 2 mol A for every 3 mol B, write the fraction so the starting unit cancels. This simple check catches many errors.

Choosing a limiting reactant by inspection: A larger mass does not necessarily mean a larger amount in moles. Different substances have different molar masses.

Confusing theoretical and actual yield: The theoretical yield is calculated; the actual yield is measured. Mixing them produces an incorrect percent yield.

Rounding too early: Keep extra digits through intermediate steps and round once at the end unless your instructor specifies another method.

Ignoring the question’s requested unit: A response in moles is not finished if the problem asks for grams, particles, or a percent.

When to revisit

Return to this checklist whenever a stoichiometry problem introduces a new type of quantity, reaction, or constraint. Before a chemistry test, practice one problem from each scenario: mole-to-mole, mass-to-mole, mass-to-mass, percent yield, and limiting reactants. If one category feels slow, review that workflow separately rather than rereading every topic.

It is also useful to revisit stoichiometry after learning related skills such as balancing equations, naming compounds, calculating molar mass, significant figures, or solution concentration. These skills are frequent prerequisites, and a weakness in one can make a stoichiometry calculation appear more difficult than it is.

Use this final pre-submission checklist:

1. Is the equation balanced?
2. Did I identify the given and requested quantities?
3. Did I convert to moles before using the mole ratio?
4. Do all units cancel correctly?
5. If there are two reactants, did I test both for the limiting reactant?
6. Did I convert to the requested final unit?
7. Is my rounding and significant-figure choice appropriate?
8. Does the result make chemical and numerical sense?

For broader preparation, pair this chemistry guide with a high school science test review plan and use the site's metric conversions practice when unit changes are slowing down your work. Consistent setup, visible units, and deliberate checking are the habits that make stoichiometry practice more reliable.