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Stoichiometric Calculations

What volume o...

Question

What volume of $H_{2} O_{(g)}$ measured at STP is produced by the combustion of $8.00 g$ of methane gas, $C H_{4}$ , according to the following equation?
$C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}$

5.60 L

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11.2 L

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22.4 L

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33.6 L

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Solution

The correct option is D $22.4 L$
Option C is the correct answer.
The equation is written as $C H_{4} + 2 O_{2}$ = $C O_{2} + 2 H_{2} O$
Here, 16 grams of $C H_{4}$ produces 36 grams of $H_{2} O$
1 gram of $C H_{4}$ produces $\frac{36}{16}$ grams of $H_{2} O$
Thus 8 grams of $C H_{4}$ produces $\frac{36}{16}$ $*$ $8$ grams of water, that is $18$ grams.
At STP. 1 mole of any gas occupies 22.4 L volume. The molecular weight of water is 18 grams. Thus 18 gram of Water will also occupy 22.4 L

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C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)

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