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2gm of Mg is burnt in a vessel containing 1.0gm of oxygen. The reactant remaining unreacted is

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0.25g of Mg

0.5g of Mg

0.1g of $O_{2}$

0.75g of Mg

answer is B.

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Detailed Solution

$M g + \frac{1}{2} O_{2} \overset{}{\to} M g O$

Limiting reagent = $O_{2}$

$16 g m O_{2} \overset{}{\to} 24 g m M g$

$1 g m O_{2} \overset{}{\to} ?$

$= \frac{24}{16} = \frac{3}{2} = 1.5 g m M g$

Wt of ‘Mg’ Left = 2.0-1.5 = 0.5gm

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