If $11.2 L$ of hydrogen and $11.2 L$ of chlorine are available for a reaction, what volume of $H C l$ will be formed? Assume STP conditions.

11.2 L

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22.4 L

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5.6 L

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44.8 L

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Solution

The correct option is B $22.4 L$
$H_{2} (g) + C l_{2} (g) \to 2 H C l (g)$
$11.2 11.2 22.4$
When $1 L i t r e$ of $H_{2}$ reacts with $1 L i t r e$ of $C l_{2}$ , it produces $2 L i t r e s$ of $H C l$ . So when $11.2 L i t r e s$ of $H_{2}$ reacts with $11.2 L i t r e s$ of $C l_{2}$ , it produces $22.4 L i t r e s$ of $H C l$ .

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