Gas Volume
$∙$ One mole hydrogen as $S T P (H_{2})$ $22.4 L$
$∙$ One mole of Nitrogen as $S T P (N_{2})$ $22.4 L$
$∙$ One mole of $C O_{2}$ at $S T P$ $22.4 L$
$∙$ One mole $H_{2} O$ at $S T P$ $22.4 L$
$∙$ One mole $N H_{3}$ at $S T P$ $22.4 L$

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Solution

$22.4 L$ of gas at $S T P = 1$ mole
$44.8 L$ of a gas at $S T P = 2$ mole
$224 L$ of a gas at $S T P = \frac{224}{22.4} = 10$ mole

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