1 g of Mg is burnt in a vessel containing 0.5 g of $O_{2} .$ The reactant remaining unreacted is

0.25 g o f M g

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0.1 g o f M g

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0.1 g o f O_{2}

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0.75 g o f M g

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Solution

The correct option is A $0.25 g o f M g$
Burning of Magnesium:
$2 M g + O_{2} \to 2 M g O$
Mass of Mg $= 1 g$
Molar mass of Mg $= 24 g$
Moles of Mg $= \frac{1}{24} = 0.041 m o l$
Mass of $O_{2} = 0.5 g$
Molar mass of $O_{2} = 32 g$
Moles of $O_{2} = \frac{0.5}{32} = 0.015 m o l$
1 mole of $O_{2}$ reacts with 2 moles of $M g$
Therefore,
O.O15 mol reacts with $2 \times 0.015 m o l$ of Mg i.e $0.03 m o l$ of Mg.
Since we have 0.041 mol of Mg, so,
$(0.041 - 0.03) \times 24 = 0.24 g$ of Mg will be left over.

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