Decomposition of Mg(NO3)₂ yields 2.5 mol gas per mol compound Steps:

• Molar mass of Mg(NO₃)₂ = 24 + 2(14) + 6(16) = 148 g/mol.
• Moles of Mg(NO₃)₂ = 0.25 / 148 ≈ 0.00169 mol.
• Reaction: Mg(NO₃)₂ → MgO + 2NO₂ + ½O₂ (2.5 mol gas/mol compound); moles gas = 0.00169 × 2.5 ≈ 0.00423 mol.
• At room conditions (≈24 L/mol = 24,000 cm³/mol), volume = 0.00423 × 24,000 ≈ 101 cm³. Why C is correct:
• Stoichiometry and ideal gas law with 24 L/mol molar volume give ≈100 cm³ total gas (NO₂ + O₂). Why the others are wrong:
• A: Underestimates gas by ignoring full NO₂ yield (≈1 mol gas/mol).
• B: Matches only NO₂ volume, ignores O₂ contribution.
• D: Overestimates by doubling gas moles (e.g., extra O₂).

Final answer: C