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Stoichiometric Calculations

12 g of Mg ...

Question

$12 g$ of $M g$ (at. mass $24$ ) will reaction completely with acid to give:

one mole of

H_{2}

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1 / 2 m o l e

H_{2}

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2 / 3 m o l e

O_{2}

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both

1 / 2 m o l e

H_{2}

and

1 / 2 m o l

O_{2}

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Solution

The correct option is A $1 / 2 m o l e$ of $H_{2}$
Let us assume the acid as $H C l$
$M g + 2 H C l ⟶ M g C l_{2} + H_{2}$

Therefore, $1$ mole of $M g$ gives $1$ mole of $H_{2}$
Moles of $M g = \frac{12}{24} = 0.5$

$∴$ Moles of $H_{2}$ produced= $\frac{1}{2}$

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

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