Limiting reagent — find the mass of product when both reagents are given

Question

10 g of hydrogen gas reacts with 64 g of oxygen gas to form water. Identify the limiting reagent and calculate the mass of water formed. Also find the mass of the excess reagent left over.

2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}

(NCERT Class 11 — fundamental stoichiometry, asked in CBSE, JEE Main, and NEET)

Solution — Step by Step

Molar mass of $\text{H}_2 = 2 \text{ g/mol}$ , molar mass of $\text{O}_2 = 32 \text{ g/mol}$ .

\text{Moles of H}_2 = \frac{10}{2} = 5 \text{ mol}

\text{Moles of O}_2 = \frac{64}{32} = 2 \text{ mol}

From the balanced equation: 2 mol $\text{H}_2$ needs 1 mol $\text{O}_2$ .

So 5 mol $\text{H}_2$ needs: $\frac{5}{2} = 2.5$ mol $\text{O}_2$ .

We only have 2 mol $\text{O}_2$ . Since we need 2.5 but have only 2, oxygen is the limiting reagent.

1 mol $\text{O}_2$ produces 2 mol $\text{H}_2\text{O}$ .

2 mol $\text{O}_2$ produces: $2 \times 2 = 4$ mol $\text{H}_2\text{O}$ .

\text{Mass of H}_2\text{O} = 4 \times 18 = \mathbf{72 \text{ g}}

2 mol $\text{O}_2$ reacts with $2 \times 2 = 4$ mol $\text{H}_2$ .

$\text{H}_2$ left over: $5 - 4 = 1$ mol $= 1 \times 2 = \mathbf{2 \text{ g}}$ .

Verify: 10 + 64 = 74 g reactants. 72 + 2 = 74 g products + leftover. Mass is conserved.

Why This Works

The balanced equation tells us the exact mole ratio in which reactants combine. If the given amounts don’t match this ratio, one reactant runs out first — that’s the limiting reagent. All product calculations must be based on the limiting reagent because once it’s consumed, the reaction stops regardless of how much of the other reactant remains.

The concept is identical to a recipe: if you need 2 eggs and 1 cup flour for each cake, and you have 5 eggs but only 2 cups flour, the flour limits you to 2 cakes. The extra egg sits unused.

Alternative Method — The Ratio Comparison

Instead of calculating how much $\text{O}_2$ is needed, compare the mole-to-coefficient ratio for each reactant:

\frac{\text{Moles of H}_2}{\text{Coefficient}} = \frac{5}{2} = 2.5

\frac{\text{Moles of O}_2}{\text{Coefficient}} = \frac{2}{1} = 2

The reactant with the smaller ratio is the limiting reagent. Here, $\text{O}_2$ has the smaller ratio (2 vs 2.5), so it’s limiting.

This ratio method is faster for JEE numerical problems with three or more reactants. Calculate the moles/coefficient ratio for each reactant — pick the smallest — that’s your limiting reagent. No need to calculate how much of each is “needed.”

Common Mistake

The classic blunder: students compare masses instead of moles. They see 64 g of $\text{O}_2$ vs 10 g of $\text{H}_2$ and conclude $\text{H}_2$ is limiting because “it has less mass.” Mass comparison is meaningless — you must convert to moles first. A small mass of $\text{H}_2$ can still represent more moles than a large mass of $\text{O}_2$ because $\text{H}_2$ has a much smaller molar mass.

Question

Solution — Step by Step

Why This Works

Alternative Method — The Ratio Comparison

Common Mistake

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