JEE Main 2025 Apr 7 Shift 2, Chemistry Q23: Stoichiometry

JEE Main2025Apr 7, Shift 2Chemistry

Butane reacts with oxygen to produce carbon dioxide and water following the equation given below

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$

If 174.0 kg of butane is mixed with 320.0 kg of O $_{2}$ , the volume of water formed in litres is _____ .(Nearest integer)

[Given : (a) Molar mass of C, H, O are 12, 1, 16 g mol $^{- 1}$ respectively, (b) Density of water = 1 g mL $^{- 1}$ ]

Your answer

Solution

Molar mass of butane, C $_{4}$ H $_{10}$ = $4 (12) + 10 (1) = 58$ g/mol, so moles of butane $= \frac{174.0 \times 1 0 ^{3}}{58} = 3 \times 1 0^{3}$ mol.

Molar mass of O $_{2}$ = 32 g/mol, so moles of O $_{2}$ $= \frac{320.0 \times 1 0 ^{3}}{32} = 10 \times 1 0^{3}$ mol.

From the stoichiometry, 1 mol butane requires 6.5 mol O $_{2}$ ; for $3 \times 1 0^{3}$ mol butane, the O $_{2}$ required is $19.5 \times 1 0^{3}$ mol, which exceeds the available $10 \times 1 0^{3}$ mol, so O $_{2}$ is the limiting reagent.

Moles of water formed $= \frac{5}{6.5} \times n_{O_{2}} = \frac{5}{13/2} \times 10 \times 1 0^{3} \approx 7.6923 \times 1 0^{3}$ mol.

Mass of water $\approx 7.6923 \times 1 0^{3} \times 18 \approx 1.3846 \times 1 0^{5}$ g. Since the density of water is 1 g/mL, this mass corresponds to $1.3846 \times 1 0^{5}$ mL $= 138.46$ litres $\approx 138$ litres.

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JEE Main 2025 Apr 7 Shift 2, Chemistry Q23: Stoichiometry

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