Butane reacts with oxygen to produce carbon dioxide and water following the equation given below…
$\mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
If 174.0 kg of butane is mixed with 320.0 kg of $\mathrm{O}_2$, the volume of water formed in litres is ________.(Nearest integer)
[Given : (a) Molar mass of $\mathrm{C}, \mathrm{H}, \mathrm{O}$ are 12, 1, $16 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively, (b) Density of water $\left.=1 \mathrm{~g} \mathrm{~mL}^{-1}\right]$
Solution
& \mathrm{C}_4 \mathrm{H}_{10}+\frac{13}{2} \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O} \\
& 3 \times 10^3 \quad 10 \times 10^3
\end{aligned}$
Moles of $\mathrm{H}_2 \mathrm{O}$ formed $=\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=5 \times \frac{2}{13} \times 10 \times 10^3$
Then $\mathrm{w}_{\mathrm{H}_2 \mathrm{O}}=\frac{10^5}{13} \times 18$
$=1.3846 \times 10^5 \mathrm{~g}$
Volume of $\mathrm{H}_2 \mathrm{O}$ will be $=138.46$ litre.
Ans. 138
Asked in: JEE Main 2025 (07 Apr Shift 2)
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