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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2025 · 7 April · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

Butane reacts with oxygen to produce carbon dioxide and water...

Butane reacts with oxygen to produce carbon dioxide and water following the equation given below
C4H10( g)+132O2( g)→4CO2( g)+5H2O(l)\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 O2\mathrm{O}_2, the volume of water formed in litres is ________.(Nearest integer)
[Given : (a) Molar mass of C,H,O\mathrm{C}, \mathrm{H}, \mathrm{O} are 12, 1, 16 g mol−116 \mathrm{~g} \mathrm{~mol}^{-1} respectively, (b) Density of water =1 g mL−1]\left.=1 \mathrm{~g} \mathrm{~mL}^{-1}\right]

Correct answer

0

Step-by-step solution

<br/>C4H10+132O2→4CO2+5H2O<br>3×10310×103<br/>\begin{aligned}<br />& \mathrm{C}_4 \mathrm{H}_{10}+\frac{13}{2} \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O} \\<br>& 3 \times 10^3 \quad 10 \times 10^3<br />\end{aligned}
Moles of H2O\mathrm{H}_2 \mathrm{O} formed =nH2O=5×213×10×103=\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=5 \times \frac{2}{13} \times 10 \times 10^3
Then wH2O=10513×18\mathrm{w}_{\mathrm{H}_2 \mathrm{O}}=\frac{10^5}{13} \times 18
=1.3846×105 g=1.3846 \times 10^5 \mathrm{~g}
Volume of H2O\mathrm{H}_2 \mathrm{O} will be =138.46=138.46 litre.
Ans. 138
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