AP Chemistry Stoichiometry- Practice Problems and Solutions

What Stoichiometry Actually Is (And Why It Matters)

Stoichiometry is the math behind chemical reactions. You use it to figure out how much of one substance reacts with another, how much product you get, and what happens when reagents aren't in perfect proportion. If you can't do stoichiometry, you can't pass AP Chemistry. Plain and simple.

Most students struggle because they memorize steps without understanding the logic. This guide fixes that. You'll get practice problems with actual solutions, broken down step by step.

The Foundation: Mole Conversions

Everything in stoichiometry starts with the mole. You need to convert between grams, moles, molecules, and liters (at STP).

Key Conversion Factors

• 1 mole = 6.022 × 10²³ particles (Avogadro's number)
• 1 mole = molar mass in grams
• 1 mole = 22.4 L at STP (for gases)

The conversion chain looks like this:

grams ⟷ moles ⟷ molecules/atoms ⟷ liters (gas)

Every stoichiometry problem is just moving between these boxes.

Balancing Chemical Equations: Non-Negotiable

You cannot solve stoichiometry problems with an unbalanced equation. It's physically impossible. The coefficients in a balanced equation give you the mole ratio—and that ratio is the entire key to every problem you'll encounter.

Example:

2H₂ + O₂ → 2H₂O

This tells you: 2 moles H₂ react with 1 mole O₂ to produce 2 moles H₂O. That's your conversion factor.

Practice Problem 1: Simple Mole Conversion

Question: How many moles are in 45.0 grams of glucose (C₆H₁₂O₆)?

Step 1: Find the molar mass.

(6 × 12.01) + (12 × 1.01) + (6 × 16.00) = 180.18 g/mol

Step 2: Use the conversion.

45.0 g × (1 mol / 180.18 g) = 0.250 mol

That's it. Convert grams to moles by dividing by molar mass. Reverse it to go moles → grams.

Practice Problem 2: Mole-to-Mole Ratio

Question: How many moles of CO₂ form when 3.5 moles of CH₄ combust?

Step 1: Write and balance the equation.

CH₄ + 2O₂ → CO₂ + 2H₂O

Step 2: Identify the mole ratio.

1 mol CH₄ : 1 mol CO₂

Step 3: Apply the ratio.

3.5 mol CH₄ × (1 mol CO₂ / 1 mol CH₄) = 3.5 mol CO₂

The mole ratio is your bridge. It comes directly from the balanced equation's coefficients.

Practice Problem 3: Grams to Grams (The Real Test)

Question: How many grams of water form when 8.0 g of H₂ reacts with excess O₂?

Step 1: Balance the equation.

2H₂ + O₂ → 2H₂O

Step 2: Convert grams H₂ → moles H₂.

8.0 g × (1 mol / 2.02 g) = 3.96 mol H₂

Step 3: Apply mole ratio.

3.96 mol H₂ × (2 mol H₂O / 2 mol H₂) = 3.96 mol H₂O

Step 4: Convert moles H₂O → grams.

3.96 mol × 18.02 g/mol = 71.4 g H₂O

This is the standard workflow: grams → moles → mole ratio → moles product → grams product.

Limiting Reagent Problems: The Hard Part

When you have multiple reactants, one runs out first. That's your limiting reagent. Everything after that is theoretical.

Question: 10.0 g of Fe reacts with 8.0 g of S. How much FeS forms?

Fe + S → FeS

Step 1: Calculate moles of each reactant.

Fe: 10.0 g × (1 mol / 55.85 g) = 0.179 mol

S: 8.0 g × (1 mol / 32.07 g) = 0.249 mol

Step 2: Determine which is limiting.

You need a 1:1 ratio. You have more moles of S than Fe. Fe runs out first. Fe is the limiting reagent.

Step 3: Calculate product from limiting reagent only.

0.179 mol Fe × (1 mol FeS / 1 mol Fe) = 0.179 mol FeS

Step 4: Convert to grams.

0.179 mol × 87.91 g/mol = 15.7 g FeS

Always base your answer on the limiting reagent. Never add theoretical yields from both reactants.

Theoretical Yield vs. Percent Yield

Theoretical yield is what you calculate on paper. Actual yield is what you measure in the lab. Real reactions don't go to completion—side reactions happen, stuff gets lost, equilibrium limits things.

Percent yield = (actual / theoretical) × 100

Question: A reaction yields 14.2 g of product in the lab, but your calculations predict 18.0 g. What's the percent yield?

(14.2 / 18.0) × 100 = 78.9%

Percent yields above 100% usually mean your product is impure or you measured wrong. Below 100% is normal.

Common Stoichiometry Mistakes

• Using unbalanced equations. This is the #1 killer. Check your coefficients every single time.
• Forgetting to convert grams to moles first. You can't use mole ratios with grams.
• Using the wrong mole ratio. The ratio comes from the balanced equation's coefficients, not your gut feeling.
• Ignoring the limiting reagent. When two reactants are given, you must identify which one runs out.
• Rounding errors. Keep extra sig figs until the final answer. Don't round intermediate steps.

Quick Reference: Stoichiometry Problem Types

Getting Started: Your Problem-Solving Workflow

1. Read the problem. Identify what you're given and what you need to find.
2. Write the balanced equation. If none is provided, write and balance it yourself.
3. Convert given amount to moles. Divide by molar mass.
4. Apply the mole ratio. Use coefficients from the balanced equation.
5. Convert moles of answer to the required unit. Grams, liters, molecules—whatever the problem asks for.
6. Check your work. Does the magnitude make sense? Did you use the right sig figs?

Work through problems in this order every time. It becomes automatic with practice.

Final Word

Stoichiometry isn't about memorization. It's about applying one consistent process to different scenarios. The mole concept, balanced equations, and unit conversions—that's the entire game. Master those three things and you can solve any stoichiometry problem on the AP exam. 🎯