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Stoichiometry practice problems: follow the mole ratio
Stoichiometry connects quantities through a balanced chemical equation. Before calculating, identify the substance you have, the substance you need and the units in the question.
A four-step method
- Balance the equation without changing chemical formulas.
- If given mass, divide by molar mass to find moles.
- Multiply by the coefficient ratio for the desired substance.
- Convert the resulting moles to the requested units.
Coefficients compare amounts in moles, not equal masses. Write units on every line so the unwanted unit cancels. For background, see OpenStax reaction stoichiometry. The exercises here are independently written.
Worked example: ammonia from nitrogen
Use N₂ + 3H₂ → 2NH₃. Suppose 2.5 mol of nitrogen reacts with enough hydrogen. How much ammonia can form?
Calculation: 2.5 mol N₂ × (2 mol NH₃ ÷ 1 mol N₂) = 5.0 mol NH₃.
The nitrogen unit cancels. The factor of two comes from the balanced equation, not from the subscript in NH₃. Subscripts identify the atoms within a molecule; coefficients describe the reaction proportions. This calculation assumes complete reaction and sufficient hydrogen.
Worked example: find the limiting reactant
Now start with 2.0 mol N₂ and 3.0 mol H₂. Calculate possible ammonia from each reactant separately:
- Nitrogen could supply 2.0 × 2 = 4.0 mol NH₃.
- Hydrogen could supply 3.0 × 2/3 = 2.0 mol NH₃.
Hydrogen limits the reaction: only 2.0 mol NH₃ can form under these assumptions. Producing that amount consumes 1.0 mol N₂, leaving 1.0 mol N₂. The smaller starting number of moles is not automatically the limiting reactant; the coefficient ratio matters.
Try these chemistry practice questions
1. For N₂ + 3H₂ → 2NH₃, how much H₂ is needed for 4.0 mol NH₃?
Show answer and steps
4.0 mol NH₃ × 3 mol H₂ / 2 mol NH₃ = 6.0 mol H₂. Check that the requested substance is in the numerator.
2. For 2Mg + O₂ → 2MgO, start with 4.0 mol Mg and 1.5 mol O₂. Identify the limiting reactant and product amount.
Show answer and steps
Mg could produce 4.0 mol MgO. O₂ could produce 3.0 mol MgO, so O₂ limits formation to 3.0 mol. There is 1.0 mol Mg left after complete reaction.
3. Using a rounded molar mass of 24.3 g/mol, convert 48.6 g Mg to moles and determine MgO formed with excess oxygen.
Show answer and steps
48.6 g ÷ 24.3 g/mol = 2.00 mol Mg. The Mg:MgO ratio is 1:1, so the theoretical product is 2.00 mol MgO. A product mass question would require another conversion using MgO's molar mass.
Review your mistake, not just the answer
If your number is wrong, mark the first incorrect step: equation, mass conversion, ratio direction or final unit. Rewrite that step with units. Then change the starting amount and solve again without the model. Keep a separate card for why an excess reactant remains; a memorized arithmetic routine does not explain the chemical limit.
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