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What volume of oxygen gas $\left(\mathrm{O}_2\right)$ measured at $0^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$, is needed to burn completely $1 \mathrm{~L}$ of propane gas $\left(\mathrm{C}_3 \mathrm{H}_8\right)$ measured under the same conditions?
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Verified Answer
The correct answer is:
$5 \mathrm{~L}$
Key Idea : Volume of a gas at $S T P=22.4 \mathrm{~L}$
$\begin{aligned}
& \mathrm{C}_3 \mathrm{H}_8+5 \mathrm{O}_2 \longrightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O} \\
& 22.4 \mathrm{~L} \quad 5 \times 22.4 \mathrm{~L}
\end{aligned}$
$\because$ To burn $22.4 \mathrm{~L} \mathrm{C}_3 \mathrm{H}_8$ the oxygen required is
$=5 \times 22.4 \mathrm{~L}$
$\therefore$ To burn $1 \mathrm{LC}_3 \mathrm{H}_8$ the oxygen required will be
$\begin{aligned}
& =\frac{5 \times 22.4}{22.4} \\
& =5 \mathrm{~L}
\end{aligned}$
$\begin{aligned}
& \mathrm{C}_3 \mathrm{H}_8+5 \mathrm{O}_2 \longrightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O} \\
& 22.4 \mathrm{~L} \quad 5 \times 22.4 \mathrm{~L}
\end{aligned}$
$\because$ To burn $22.4 \mathrm{~L} \mathrm{C}_3 \mathrm{H}_8$ the oxygen required is
$=5 \times 22.4 \mathrm{~L}$
$\therefore$ To burn $1 \mathrm{LC}_3 \mathrm{H}_8$ the oxygen required will be
$\begin{aligned}
& =\frac{5 \times 22.4}{22.4} \\
& =5 \mathrm{~L}
\end{aligned}$
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