JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Some Basic Concepts of Chemistry

The molar mass of butane is $58$ g/mol, so $174.0$ kg gives $3000$ mol, while $320.0$ kg of oxygen gives $10000$ mol. The equation needs $6.5$ mol of oxygen for each mole of butane, which would demand $19500$ mol of oxygen to burn all the butane; since only $10000$ mol is present, oxygen is the limiting reagent. As $6.5$ mol of oxygen yields $5$ mol of water, the $10000$ mol of oxygen gives $10000\times \frac{5}{6.5}=\frac{100000}{13}$ mol of water. This is a mass of $\frac{100000}{13}\times 18\approx 1.385\times 10^5$ g, and with a density of $1$ g/mL the volume of water formed is about $138$ litres.