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1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24; O = 16)

1. Mg, 0.16 g
2. O2, 0.16 g
3. Mg, 0.44 g
4. O2, 0.28 g

Correct answer: Mg, 0.16 g

Solution

The reaction is 2Mg + O2 → 2MgO. Moles of Mg = 1.0/24 = 0.0417, and moles of O2 = 0.56/32 = 0.0175. Mg requires 0.0175 × 2 = 0.035 moles, leaving 0.0417 - 0.035 = 0.0067 moles of Mg unreacted, which is 0.0067 × 24 = 0.16 g.

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