Master the 3-step process: find the limiting reactant, calculate theoretical yield, then compare to actual yield for percent yield.
This worksheet tests three connected skills. Every problem uses the same framework.
To go from grams of reactant → grams of product, always go through moles of reactant, apply the balanced equation's mole ratio, then convert back to grams.
Convert both reactants from grams to moles using their molar masses.
Divide each reactant's moles by its coefficient. The smaller value is limiting.
From the limiting reactant, use the balanced equation's mole ratio to find moles of product.
Multiply moles of product by its molar mass to get the theoretical yield in grams.
Divide actual yield by theoretical yield and multiply by 100%.
Actual yield comes from the experiment. Theoretical yield comes from your calculations (step 4 above).
Molar mass comes from adding up atomic masses on the periodic table (in g/mol).
The limiting reactant is the one that runs out first. It controls how much product you can actually make.
If you have 10 slices of bread and 3 slices of cheese, you can only make 3 sandwiches — the cheese runs out first. The cheese is your limiting reactant. The extra bread is called "excess."
In chemistry, it's the same — one reactant always runs out before the others, and that limits how much product you can form.
After converting both reactants to moles:
Problem: 4 mol of hydrogen reacts with 3 mol of oxygen. Which is limiting? 2H₂ + O₂ → 2H₂O
Theoretical yield is the maximum amount of product you could make if everything went perfectly — based on the limiting reactant.
Starting from the limiting reactant's moles:
Problem: Using 2.0 mol hydrogen (limiting) in 2H₂ + O₂ → 2H₂O, how many grams of water are produced?
Real reactions never reach 100% — some product is lost or never forms. Percent yield tells you how efficient the reaction was.
Problem: The theoretical yield is 36.0 g of water, but you actually collected 29.5 g. What's the percent yield?
Use these to check your work or practice the calculations.
Input both reactants and the product, get the limiting reactant and theoretical yield.
The ones you'll need for this worksheet:
Each problem has multiple parts. Fill in all parts, then "Check All Answers" for per-part feedback. "Show Solution" reveals full step-by-step work.
Grams ÷ molar mass → Moles
Moles × mole ratio → Moles (new)
Moles × molar mass → Grams
Always go through moles.
1. Convert both reactants to moles
2. Divide each by its coefficient
3. Smaller value = LR
4. Use LR's moles in mole ratio
5. Convert product moles to grams
% Yield = (actual ÷ theoretical) × 100%
Actual = experimental result
Theoretical = what stoichiometry predicts
Should be ≤ 100%
H = 1.008 · C = 12.01
N = 14.01 · O = 16.00
F = 19.00 · Na = 22.99
P = 30.97 · S = 32.07
Cl = 35.45 · K = 39.10
Br = 79.90 · I = 126.90
• Using non-limiting reactant for yield
• Flipping % yield (theoretical/actual)
• Forgetting to divide moles by coefficient
• Rounding too early in the calculation
• Wrong mole ratio from unbalanced eq.
Avogadro's # = 6.022 × 10²³ /mol
Molar volume at STP = 22.4 L/mol
1 mol = 1 molar mass in grams
Balanced equation is non-negotiable!
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