What volume of $H_{2} O$ is required to produce $5$ L $O_{2}$ by the following equation: ${2 H}_{2} O (g) \to {2 H}_{2} (g) + O_{2} (g)$

3

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5

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10

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16

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14

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Solution

The correct option is C $10$ L
The unbalanced chemical equation is $H_{2} O (g) \to H_{2} (g) + O_{2} (g)$ .
The balanced chemical equation is $2 H_{2} O (g) \to 2 H_{2} (g) + O_{2} (g)$
Thus 1 mole of $O_{2}$ is obtained from 2 moles of $H_{2} O$ .
Also, the volume is directly proportional to the number of moles.
Hence, $5$ L of $O_{2}$ will be obtained from $5 \times \frac{2}{1} = 10$ L of $H_{2} O$ .

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