Stoichiometry Examples- Practice Problems and Solutions

What Stoichiometry Actually Is

Stoichiometry is just math applied to chemical reactions. You take a balanced equation and use it to calculate how much of one substance you need or will get from another. That's it. No magic, no hidden tricks—just proportions and arithmetic.

Most students struggle because they skip the fundamentals and jump straight into complex problems. If you don't know what a mole is or can't balance an equation in your sleep, stop here. Go back. Learn those first.

The Core Concepts You Need

Before touching any problem, these must be automatic:

• Mole: 6.022 × 10²³ particles. It's a counting unit, like a dozen.
• Molar mass: Grams per mole. You get this from the periodic table—add up the atomic masses.
• Balanced equation: Same number of atoms on both sides. No exceptions.
• Mole ratio: The coefficients in your balanced equation. These are your conversion factors.

If any of these make you pause, you're not ready for the problems below. Go study first.

The Basic Process: How to Actually Solve These

Every stoichiometry problem follows the same roadmap:

1. Balance your equation (if it isn't already)
2. Convert what you're given to moles
3. Use the mole ratio to find moles of the unknown
4. Convert back to grams, liters, or whatever units you need

That's the entire process. Memorize it. This is your algorithm.

Example 1: Mass-to-Mass Conversion

Problem: How many grams of water form when 4 grams of hydrogen react with excess oxygen?

Step 1: Write and balance the equation

2H₂ + O₂ → 2H₂O

Step 2: Convert grams of H₂ to moles

4 g H₂ × (1 mol H₂ / 2 g H₂) = 2 mol H₂

Step 3: Use mole ratio to find moles of H₂O

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

Step 4: Convert moles of H₂O to grams

2 mol H₂O × (18 g H₂O / 1 mol H₂O) = 36 g H₂O

That's your answer. Each step follows the dimensional analysis method—units cancel until you're left with what you need.

Example 2: Limiting Reagent Problem

Limiting reagents trip people up. Here's how to handle them.

Problem: 10 g of Fe reacts with 8 g of S. How much FeS forms?

Step 1: Balance the equation

Fe + S → FeS

Step 2: Calculate how much product each reactant could make

For Fe: 10 g × (1 mol / 56 g) × (1 mol FeS / 1 mol Fe) × (88 g / 1 mol) = 15.7 g FeS

For S: 8 g × (1 mol / 32 g) × (1 mol FeS / 1 mol S) × (88 g / 1 mol) = 22 g FeS

Step 3: The smaller answer wins

Fe makes less product, so Fe is the limiting reagent. Your answer is 15.7 g FeS.

Never add the answers together. The limiting reagent determines the maximum product.

Example 3: Volume-Based Calculation (Gas Stoichiometry)

At standard temperature and pressure, 1 mole of any gas occupies 22.4 L.

Problem: How many liters of CO₂ form when 5 L of propane (C₃H₈) burn completely?

Step 1: Balance the combustion equation

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Step 2: Apply the mole ratio

5 L C₃H₈ × (3 L CO₂ / 1 L C₃H₈) = 15 L CO₂

When gases are at the same temperature and pressure, the volume ratio equals the mole ratio. You don't even need to convert to moles first.

Percent Yield: When Reactions Don't Go to Completion

Theoretical yield is what the math says should happen. Actual yield is what you actually get in the lab. Percent yield compares them.

Formula: (Actual / Theoretical) × 100

Example: You calculate 50 g of product should form. You actually get 35 g. Percent yield = (35/50) × 100 = 70%.

If your percent yield is over 100%, you did something wrong. Contamination, measurement errors, or incomplete drying. It happens. Check your work.

Common Mistakes That Will Sink You

• Forgetting to balance the equation. This invalidates everything that follows.
• Using atomic mass wrong. Round to whole numbers for simplicity, but be consistent.
• Forgetting to convert to moles first. You can't do stoichiometry in grams directly.
• Using the wrong mole ratio. Check your coefficients twice.
• Not identifying the limiting reagent. If you have amounts of multiple reactants, you must check which one runs out first.

Practice Problems

Try these. No peeking at solutions until you've tried.

Problem 1: How many grams of NaCl form when 23 g of Na react with excess Cl₂?

2Na + Cl₂ → 2NaCl

Problem 2: 15 g of N₂ reacts with 15 g of H₂. Which reagent limits the reaction? How much NH₃ forms?

N₂ + 3H₂ → 2NH₃

Problem 3: If 40 g of a product has a percent yield of 75%, what was the theoretical yield?

Solutions to Practice Problems

Solution 1:

23 g Na × (1 mol / 23 g) × (2 mol NaCl / 2 mol Na) × (58.5 g / 1 mol) = 58.5 g NaCl

Solution 2:

N₂: 15 g × (1 mol / 28 g) × (2 mol NH₃ / 1 mol N₂) × (17 g / 1 mol) = 18.2 g NH₃

H₂: 15 g × (1 mol / 2 g) × (2 mol NH₃ / 3 mol H₂) × (17 g / 1 mol) = 8.5 g NH₃

H₂ is limiting. Answer: 8.5 g NH₃

Solution 3:

40 = (Actual / Theoretical) × 100

Theoretical = 40 / 0.75 = 53.3 g

Quick Reference: Stoichiometry at a Glance

The Bottom Line

Stoichiometry is procedural. Memorize the steps. Practice until you can do them without thinking. Use dimensional analysis—it's foolproof if you set it up right.

Get the basics down: moles, molar mass, balanced equations, mole ratios. Everything else builds from there. Skip the fundamentals and you'll keep failing the same way.

Do problems. More than you think you need. That's the only way this stuff sticks.