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1.0 g of magnesium is burnt with 0.56 g $O_{2}$ in a closed vessel. Which reactant is left in excess and how much?
(At. wt. Mg = 24,O = 16 )

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Mg, 0.16g

$O_{2}, 0.16 g$

Mg, 0.44g

$O_{2}, 0.28 g$

answer is A.

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Detailed Solution

$n_{Mg} = \frac{1}{24} = 0.0416 moles$

$n_{O_{2}} = \frac{0.56}{32} = 0.0175 moles$

the balanced equation is

$∴ Mass of Mg left in excess = 0.0066 \times 24 = 0.16 g$

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