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Question

Volume of oxygen required for complete combustion of 12.5 $$dm^{3}$$ of benzene at NTP is:


A
87.5 dm3
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B
43.75 dm3
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C
40 dm3
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D
80 dm3
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Solution

The correct option is A 87.5 $$dm^{3}$$
The chemical reaction can be represented as:

$$C_6H_6+7.5O_2\rightarrow 6CO_2+3H_2O$$

Since at NTP, 1 mole=22.4 litres, hence

According to the reaction, $$22.4\ L$$ benzene requires $$7.5\times 22.4=168\ L$$ oxygen for complete combustion

Therefore, $$12.5\ dm^3$$ benzene will require $$\dfrac{168}{22.4}\times12.5=87.5\ dm^3$$ oxygen for complete combustion

Hence, option $$A$$ is the correct answer.

Chemistry

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