The minimum amount of O2(g) consumed per gram of reactant is for the reaction: (Given atomic mass: Fe=56, O=16, Mg=24, P=31, C=12, H=1)

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
P4(s) + 5O2(g) → P4O10(s)
4Fe(s) + 3O2(g) → 2Fe2O3(s)
2Mg(s) + O2(g) → 2MgO(s)

The molar mass of C3H8 = 3(12) + 8(1) = 44 g/mol
The molar mass of O2 = 2(16) = 32 g/mol
The molar mass of P4 = 4(31) = 124 g/mol
The molar mass of Fe = 56 g/mol
The molar mass of Mg = 24 g/mol

For C3H8 + 5O2 → 3CO2 + 4H2O
44 g of C3H8 reacts with 5 * 32 = 160 g of O2
1 g of C3H8 reacts with 160/44 = 3.64 g of O2

For P4 + 5O2 → P4O10
124 g of P4 reacts with 5 * 32 = 160 g of O2
1 g of P4 reacts with 160/124 = 1.29 g of O2

For 4Fe + 3O2 → 2Fe2O3
4 * 56 = 224 g of Fe reacts with 3 * 32 = 96 g of O2
1 g of Fe reacts with 96/224 = 0.43 g of O2

For 2Mg + O2 → 2MgO
2 * 24 = 48 g of Mg reacts with 32 g of O2
1 g of Mg reacts with 32/48 = 0.67 g of O2

The minimum amount of O2 consumed per gram of reactant is 0.43 g for the reaction 4Fe(s) + 3O2(g) → 2Fe2O3(s)