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WAEC Chemistry 2020 Paper

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Question : 30

Total: 49

What volume of oxygen at s.t.p is required to burn completely

7.5\ \mathrm{dm}^3

of methane according to the following equation?

\mathrm{CH}_4\mathrm{(g)} + 2\,\mathrm{O}_2\mathrm{(g)} \rightarrow \mathrm{CO}_2\mathrm{(g)} + \mathrm{H}_2\mathrm{O}\mathrm{(g)}

$3.75\ \mathrm{dm}^3$
$7.59\ \mathrm{dm}^3$
$\;15.0\ \mathrm{dm}^3$
$\;30.0\ \mathrm{dm}^3$

Solution:

Concept:

Gay-Lussac's law states that gases react in simple whole-number volume ratios at the same temperature and pressure.

Explanation:

From the balanced equation,

\mathrm{CH}_4\mathrm{(g)} + 2\,\mathrm{O}_2\mathrm{(g)} \rightarrow \mathrm{CO}_2\mathrm{(g)} + 2\,\mathrm{H}_2\mathrm{O}\mathrm{(g)}

,

1 volume of methane reacts with 2 volumes of oxygen.

Thus, the volume ratio of

\mathrm{CH}_4 : \mathrm{O}_2

is

1 : 2

.

For

7.5\ \mathrm{dm}^3

of methane,

volume of oxygen required

= 7.5 \times 2 = 15.0\ \mathrm{dm}^3

.

Therefore,

15.0\ \mathrm{dm}^3

of oxygen is needed.

Answer:

C.

15.0\ \mathrm{dm}^3

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