What volume of hydrogen is required to react with 31 L of oxygen under the same conditions? 2 H2(g) + O2(g) → 2 H2O(g)

62L

Explanation:

2 H2(g) + O2(g) → 2 H2O(g)

From the stoichiometry of the balanced equation;

2 mol of H2 reacts with 1 mol of O2.

From the law of combining volumes, equal moles of gases occupies equal volume.

This means;

2 L of H2 reacts with 1 L of O2

x L of hydrogen reacts with 31 L of O2

2 = 1

x = 31

Solving for x;

x = 62 L

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