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The minimum amount of O 2 (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)

1. 4 Fe(s) + 3 O 2 (g) → 2 Fe 2 O 3 (s)
2. P 4 (s) + 5 O 2 (g) → P 4 O 10 (s)
3. C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(l)
4. 2Mg(s) + O 2 (g) → MgO(s)

Correct answer: 4 Fe(s) + 3 O 2 (g) → 2 Fe 2 O 3 (s)

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