Question

Consider the following chemical equation. $$ 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ If \(25.0 \mathrm{~mL} \mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

Short answer

The volume of N₂O₄ gas produced when 25.0 mL of NO₂ gas is completely converted under the same conditions is \(12.5 \mathrm{~mL}\).

Step-by-step solution

Write down the balanced chemical equation

The balanced chemical equation is given by: \[ 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) \]

Identify the stoichiometry of the reaction

According to the balanced chemical equation, 2 moles of NO₂ combine to form 1 mole of N₂O₄. We can express this stoichiometry as a ratio: \[ \frac{\text{moles of NO}_2}{\text{moles of N}_2 \text{O}_4} = \frac{2}{1} \]

Use the stoichiometry to find the volume of N₂O₄

Since the temperature and pressure conditions remain the same, we can apply the stoichiometry of the reaction directly to the volumes: \[ \frac{\text{volume of NO}_2}{\text{volume of N}_2 \text{O}_4} = \frac{2}{1} \] We are given the volume of NO₂ gas as \( 25.0 \mathrm{~mL} \). We can now solve for the volume of N₂O₄ gas: \[ \text{volume of N}_2 \text{O}_4 = \frac{1}{2} \times \text{volume of NO}_2 \]

Calculate the volume of N₂O₄

Substitute the given volume of NO₂ gas and calculate the volume of N₂O₄ gas: \[ \text{volume of N}_2 \text{O}_4 = \frac{1}{2} \times 25.0 \mathrm{~mL} = 12.5 \mathrm{~mL} \] Hence, the volume of N₂O₄ gas produced when 25.0 mL of NO₂ gas is completely converted is 12.5 mL