What is the change in enthalpy when 11.8g of iron reacts with 2.7 L of O2 (at STP) according to the following equation?

Question

What is the change in enthalpy when 11.8g of iron reacts with 2.7 L of O₂ (at STP) according to the following equation? Hint: 1 mol of any gas at STP = 22.4 L

3Fe + 2O₂  → Fe₃O₄    ∆H^o = -1120.48kJ

Answer 1

mols Fe = 11.8 g x 1 mol / 55.8 g = 0.211 mols Fe

mols O₂ = 2.7 L x 1 mol / 22.4 L = 0.121 mols O₂

Based on the mol ratio of 3:2 for Fe:O₂, O₂ is LIMITING

mols Fe₃O₄ formed = 0.121 mols O2 x 1 mol Fe₃O₄ / 2 mols O2 = 0.0605 mols

∆H = 0.0605 mols Fe₃O₄ x -1120.48 kJ / mol = 67.8 kJ = 68 kJ (2 sig. figs.)