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$12 g$ of Mg (At. mass 24) will react completely with acid to give

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$1 / 2 mole of H_{2}$

$2 / 3 mole of O_{2}$

Both $1 / 2 mol$ of $H_{2}$ and $1 / 2 mol of O_{2}$

One mole of $H_{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} {Mg}^{+ 2} \equiv H_{2} \\ n = \frac{12 gm}{24 gm} = \frac{1}{2} mole of H_{2} \end{array}$

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