SCH3U Stoichiometry Practice: Moles, Ratios and Limiting Reactants

To solve a stoichiometry problem, balance the chemical equation, convert the given quantity to moles, use the coefficient ratio, and convert to the requested unit. When quantities of two reactants are supplied, check which one limits the reaction before calculating the product.

This original SCH3U practice set focuses on reasoning through those steps. It is not an official assessment. Use the supplied molar masses so your answers can be checked consistently, and assume complete reaction where the question states it.

Why do different solution methods look different?

An Ontario student discussion asks how much stoichiometry to review when online methods look unfamiliar. The useful response is to track units and mole ratios: a proportion and a dimensional-analysis chain can express the same relationship. Do not judge a method only by its layout. Student question

For example, n = m/M and m = nM are rearrangements of the same relationship. The balanced equation supplies a separate relationship between substances. Confusing those two roles is a common source of incorrect conversions.

A four-step setup you can reuse

• 1. Write and balance the equation without changing chemical subscripts.
• 2. Convert each relevant given amount to moles.
• 3. Apply the coefficient ratio or identify the limiting reactant.
• 4. Convert the result to the requested quantity, retaining units.

Try the eight questions before reading the explanations.

Eight practice questions

• 1. Balance H₂ + O₂ → H₂O using the smallest whole-number coefficients.
• 2. How many moles are in 9.00 g H₂O? Use M(H₂O) = 18.0 g/mol.
• 3. For 2H₂ + O₂ → 2H₂O, what amount of water forms from 0.400 mol O₂ with excess H₂?
• 4. CaCO₃ → CaO + CO₂. Find the mass of CO₂ formed by complete decomposition of 10.0 g CaCO₃. Use molar masses 100.0 and 44.0 g/mol respectively.
• 5. For N₂ + 3H₂ → 2NH₃, identify the limiting reactant when 2.00 mol N₂ and 3.00 mol H₂ are available. Find the amount of NH₃ formed.
• 6. In Question 5, how much N₂ remains after the limiting reactant is consumed?
• 7. An experiment produces 3.52 g product when the theoretical yield is 4.40 g. Calculate percent yield.
• 8. HCl + NaOH → NaCl + H₂O. What volume of 0.100 mol/L NaOH is needed to react with 25.0 mL of 0.200 mol/L HCl?

Worked answers

1. Balance atoms, not the formula

The balanced equation is 2H₂ + O₂ → 2H₂O. There are four H atoms and two O atoms on each side. Changing H₂O to H₂O₂ would change the substance; balancing uses coefficients instead.

2. Convert grams to moles

n = 9.00 g ÷ 18.0 g/mol = 0.500 mol. Dividing mass by molar mass leaves moles. Keep the supplied values through the calculation and round the result appropriately.

3. Use the coefficient ratio

One mole of O₂ corresponds to two moles of H₂O. Therefore, 0.400 mol O₂ × (2 mol H₂O / 1 mol O₂) = 0.800 mol H₂O. Excess H₂ means it does not limit this calculation.

4. Follow the mass–mole–mass route

First, n(CaCO₃) = 10.0/100.0 = 0.100 mol. The equation gives a 1:1 ratio, so n(CO₂) = 0.100 mol. Then m(CO₂) = 0.100 × 44.0 = 4.40 g. A 1:1 mole ratio does not mean equal reactant and product masses.

5. Compare product amounts from each reactant

The N₂ could produce 2.00 × 2 = 4.00 mol NH₃ if sufficient H₂ were available. The H₂ can produce 3.00 × 2/3 = 2.00 mol NH₃. The smaller possible product amount controls the result: H₂ is limiting and 2.00 mol NH₃ forms.

Do not choose the limiting reactant simply by finding the smaller starting mass or mole amount. The required ratio matters.

6. Find the amount consumed before subtracting

Consuming 3.00 mol H₂ requires 1.00 mol N₂. The remaining N₂ is 2.00 − 1.00 = 1.00 mol. The limiting reactant is used up in this idealized complete-reaction model; the excess reactant is not necessarily used up.

7. Compare actual with theoretical yield

Percent yield = actual/theoretical × 100 = 3.52/4.40 × 100 = 80.0%. Reversing the fraction would not answer the question. An apparent yield above 100% calls for checking measurements, product purity and the calculation, rather than assuming more matter was created.

8. Convert volume and apply the 1:1 ratio

25.0 mL = 0.0250 L. The amount of HCl is 0.200 × 0.0250 = 0.00500 mol. The equation requires 0.00500 mol NaOH. Its volume is V = n/c = 0.00500/0.100 = 0.0500 L = 50.0 mL.

What to practise after checking your answers

If Questions 2 and 4 were difficult, review units and molar mass. If Questions 3 and 5 were difficult, write the coefficient ratio explicitly. If Question 8 was difficult, separate the concentration relationship from the reaction relationship before combining them.

Connect practice to your course

See the SCH3U guide and Grade 11 science tutoring. Book a free one-hour assessment for a personalised starting point.

Curriculum reference: Ontario Grades 11–12 Science, SCH3U quantities in chemical reactions.