JEE Main 2025 — Some Basic Concepts Of Chemistry Question with Solution

The chemical reaction between butane (${\mathrm{C}}_4{\mathrm{H}}_{10}$) and oxygen (${\mathrm{O}}_2$) to produce carbon dioxide (${\mathrm{CO}}_2$) and water (${\mathrm{H}}_2\mathrm{O}$) can be represented by the following equation:

${\mathrm{C}}_4{\mathrm{H}}_{10}(\mathrm{g})+\frac{13}{2}{\mathrm{O}}_2(\mathrm{g})\to 4{\mathrm{CO}}_2(\mathrm{g})+5{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Given that we have 174.0 kg of butane and 320.0 kg of oxygen, we are tasked with calculating the volume of water produced in liters.

Mole Calculation for Water:

The stoichiometry of the reaction shows that 5 moles of water are produced per 13/2 moles of oxygen.

Using the moles of oxygen available from input, calculate the moles of water produced:

$\text{Moles of }{\mathrm{H}}_2\mathrm{O}=5\times \frac{2}{13}\times (10\times 10^3)$

Mass of Water Produced:

Calculate the mass of water using the molar mass of water (18 g/mol):

$\text{Mass of }{\mathrm{H}}_2\mathrm{O}=\left(\frac{10^5}{13}\right)\times 18$

The result will be:

$=1.3846\times 10^5\text{g}$

Volume of Water:

Convert the mass of water to volume, considering the density of water is $1\text{g/mL}$ (or $1\text{g/L}$):

$\text{Volume of }{\mathrm{H}}_2\mathrm{O}=138.46\text{liters}$

Therefore, the volume of water formed is approximately 138 liters when rounded to the nearest integer.