Question

what volume, in l, of water vapor will be produced when 22.0 l of h₂ and 11.0 l of o₂ react according to the equation 2 h₂(g) + o₂(g) → 2 h₂o(g) assume that the reactants and product are all at the same temperature and pressure.

Explanation:

Step1: Identify Limiting Reactant

From the reaction \( 2\text{H}_2(\text{g}) + \text{O}_2(\text{g})
ightarrow 2\text{H}_2\text{O}(\text{g}) \), the volume ratio of \( \text{H}_2 \) to \( \text{O}_2 \) is \( 2:1 \). Given \( V(\text{H}_2) = 22.0 \, \text{L} \) and \( V(\text{O}_2) = 11.0 \, \text{L} \), the ratio of \( \text{H}_2 \) to \( \text{O}_2 \) is \( \frac{22.0}{11.0} = 2 \), which matches the stoichiometric ratio. So both reactants are fully consumed.

Step2: Relate Volume of \( \text{H}_2 \) to \( \text{H}_2\text{O} \)

The volume ratio of \( \text{H}_2 \) to \( \text{H}_2\text{O} \) is \( 2:2 = 1:1 \) (from the balanced equation). So \( V(\text{H}_2\text{O}) = V(\text{H}_2) = 22.0 \, \text{L} \) (or using \( \text{O}_2 \): ratio \( \text{O}_2 \) to \( \text{H}_2\text{O} \) is \( 1:2 \), so \( V(\text{H}_2\text{O}) = 2 \times V(\text{O}_2) = 2 \times 11.0 = 22.0 \, \text{L} \)).

Answer:

\( 22.0 \, \text{L} \)