How many litres of oxygen are required for complete combustion of 39 grams of liquid benzene at STP?

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Solution

The correct option is A 84
The combustion reaction is:

$C_{6} H_{6} + \frac{15}{2} O_{2} \to 6 C O_{2} + 3 H_{2} O$

This means that $\frac{15 \times 22.4}{2}$ L of $O_{2}$ at STP is required for $78$ grams of benzene. (Mol. wt of Benzene $=$ $78$ )

For $39$ grams of benzene, $\frac{39 \times 15 \times 22.4}{78 \times 2}$ $=$ $84$ grams of $O_{2}$ is required.

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Similar questions

2 C_{6} H_{6} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O

39

C_{6} H_{6}

S T P

39 g m s

C = 12, H = 1, O = 16

(C_{6} H_{6})

2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)

39 g

C_{6} H_{6}

39

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