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$\mathrm{C}_6 \mathrm{H}_6(\mathrm{liq})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{liq})$
Benzene burns in oxygen according to the above equation. What is the volume of oxygen (at STP) needed for complete combustion of 39 gram of liquid benzene?
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Benzene burns in oxygen according to the above equation. What is the volume of oxygen (at STP) needed for complete combustion of 39 gram of liquid benzene?
Solution:
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Verified Answer
The correct answer is:
84 litre
$\mathrm{n}_{\mathrm{C}_6 \mathrm{H}_6}=\frac{39}{78}=\frac{1}{2}$ mole
From Balanced equation,
$\begin{aligned}
&\mathrm{n}_{\mathrm{C}_6 \mathrm{H}_{\mathrm{G}}}=\frac{2}{15} \mathrm{n}_{\mathrm{O}_2} \\
&\frac{1}{2}=\frac{2}{15} \times \frac{\mathrm{xL}}{22.4 \mathrm{~L}} \\
&\therefore \mathrm{x}=84 \mathrm{~L}
\end{aligned}$
From Balanced equation,
$\begin{aligned}
&\mathrm{n}_{\mathrm{C}_6 \mathrm{H}_{\mathrm{G}}}=\frac{2}{15} \mathrm{n}_{\mathrm{O}_2} \\
&\frac{1}{2}=\frac{2}{15} \times \frac{\mathrm{xL}}{22.4 \mathrm{~L}} \\
&\therefore \mathrm{x}=84 \mathrm{~L}
\end{aligned}$
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