One of the reactions in the petroleum industry for improving the octane rating of fuel is $C_{7} H_{14} \to C_{7} H_{8} + 3 H_{2}$ . The two hydrocarbons in this equation are liquid while $H_{2}$ is a gas. What is the per cent reduction in liquid weight accompanying the above reaction?

9.39 %

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0.939 %

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6.1 %

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0.61 %

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Solution

The correct option is C $6.1 %$
The balanced reaction is as follows:

$C_{7} H_{14} (l) \to C_{7} H_{8} (l) + 3 H_{2} (g)$

$98$ g $92$ g $6$ g

Reduction in the weight of liquid $= 98 - 92 = 6$ g

Percentage reduction $=$ $\frac{6}{98} \times 100 = 6.1 %$

Hence, option C is correct.

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