A gas mixture of 3 litres of propane $\left(\mathrm{C}_{3} \mathrm{H}_{8}ight)$ and butane…
- $2: 1$
- $1: 2$
- $1.5: 1.5$
- $0.5: 2.5$
Solution
\mathrm{C}_{3} \mathrm{H}_{8}+5 \mathrm{O}_{2} ightarrow 3 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O} \\
\mathrm{a} \qquad \qquad \qquad \quad 3 \mathrm{a}
\end{array}$
$\begin{array}{ll}
\mathrm{C}_{4} \mathrm{H}_{10}+\frac{13}{2} \mathrm{O}_{2} ightarrow 4 \mathrm{CO}_{2}+5 \mathrm{H}_{2} \mathrm{O} \\
(3-a) \qquad \quad \quad 4(3-a)
\end{array}$
But, $3 \mathrm{a}+4(3-\mathrm{a})=10$
$a=2$ (Propane) and $3-2=1$ (Butane) ~
Asked in: JEE-TOPICTESTS-CHEMISTRY
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