What quantity of oxygen is required for complete burning of $15.6 \mathrm{~g}$ benzene?
What quantity of oxygen is required for complete burning of $15.6 \mathrm{~g}$ benzene?
- $75 \mathrm{~g}$
- $88 \mathrm{~g}$
- $48 \mathrm{~g}$
- $64 \mathrm{~g}$
Solution
$\begin{aligned} & \mathrm{C}_6 \mathrm{H}_6+\frac{15}{2} \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O} \\ & \frac{15.6}{78}=0.2 \mathrm{~mol} \\ & 1.5 \mathrm{~mol} \times 32 \\ & =48 \mathrm{~g}\end{aligned}$
Asked in: MHT CET 2022 (07 Aug Shift 1)
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