The minimum amount of O2(g) consumed per gram of reactant is for the reaction: (Atomic mass: Fe = 56, O = 16, Mg = 24,…

🧠 Per gram of reactant — not per mole The question asks for the smallest mass of $\ce{O2}$ needed per gram of the other reactant. Heavy metals like Fe lose few electrons per gram, so they consume very little oxygen per gram. Light fuels like propane burn a lot of oxygen per gram.

🗺️ The four ratios (mass $\ce{O2}$ per $1\,\text{g}$ reactant) (a) $\ce{P4}$: $(5 \times 32) / (4 \times 31) = 160/124 \approx 1.29\,\text{g}$. (b) $\ce{Mg}$: $(1 \times 32) / (2 \times 24) = 32/48 \approx 0.67\,\text{g}$. (c) $\ce{Fe}$: $(3 \times 32) / (4 \times 56) = 96/224 \approx 0.43\,\text{g}$. (d) $\ce{C3H8}$: $(5 \times 32) / 44 \approx 3.64\,\text{g}$.

The smallest is $0.43\,\text{g}$ for iron.

⚠️ Common mistake Picking propane because "$5\,\ce{O2}$" looks small. The mole count is per mole of fuel, but the question wants per gram. Iron is heavy, so each gram contains few atoms — so few oxygens are needed.

$\boxed{\text{Answer: (c)}}$