To determine the volume of gas produced when 0.1 moles of magnesium nitrate (Mg(NO₃)₂) decomposes completely, we need to use the stoichiometry of the given chemical equation and the ideal gas law's approximation that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 dm³.
1. Balanced Chemical Equation:
[tex]\[
2 \, Mg(NO_3)_2(s) \rightarrow 2 \, MgO(s) + 4 \, NO_2(g) + O_2(g)
\][/tex]
2. Determine the moles of gases produced:
- According to the balanced equation, 2 moles of Mg(NO₃)₂ produce 4 moles of NO₂ and 1 mole of O₂.
- Therefore, 1 mole of Mg(NO₃)₂ produces 2 moles of NO₂ and 0.5 mole of O₂.
3. For 0.1 moles of Mg(NO₃)₂:
- Moles of NO₂ produced: [tex]\(0.1 \, \text{moles of Mg(NO₃)₂} \times 2 \, \frac{\text{moles of NO₂}}{\text{moles of Mg(NO₃)₂}} = 0.2 \, \text{moles of NO₂}\)[/tex]
- Moles of O₂ produced: [tex]\(0.1 \, \text{moles of Mg(NO₃)₂} \times 0.5 \, \frac{\text{moles of O₂}}{\text{moles of Mg(NO₃)₂}} = 0.05 \, \text{moles of O₂}\)[/tex]
4. Total moles of gas produced:
[tex]\[
0.2 \, \text{moles of NO₂} + 0.05 \, \text{moles of O₂} = 0.25 \, \text{moles of gas}
\][/tex]
5. Volume of gas at STP:
- Given that 1 mole of any gas at STP occupies 22.4 dm³:
[tex]\[
\text{Volume of gas} = 0.25 \, \text{moles} \times 22.4 \, \frac{\text{dm}^3}{\text{mole}} = 5.6 \, \text{dm}^3
\][/tex]
Therefore, the volume of gas produced when 0.1 moles of magnesium nitrate decompose completely is:
[tex]\[
\boxed{5.6 \, \text{dm}^3}
\][/tex]
Since none of the provided options match this volume exactly, it seems there was a mistake in the provided options. The correct volume should be [tex]\(5.6 \, \text{dm}^3\)[/tex].
