Chemistry Q71 – JEE-MAIN 2025

Home Chemistry Basic ChemistryQuestion #76

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)

174.0

kg of butane is mixed with

320.0

kg of

O_{2}

, the volume of water formed in liters is _______. (Nearest integer) [Given : (a) Molar mass of C, H, O are

12, 1, 16

m o l^{- 1}

respectively, (b) Density of water =

1

m L^{- 1}

]

Enter Calculated Numerical Value:

Clue 1 of 3: Stoichiometry and Limiting Reagent

In a chemical reaction with multiple reactants, the amount of product formed is determined by the limiting reagent, which is the reactant that is completely consumed first.

🧠 Full Solution Path

Step 1: Balance the chemical equation and calculate initial moles.✦ Active

First, ensure the chemical equation is balanced. The given equation is $C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$ . To work with whole numbers, multiply by 2: $2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{2} O (l)$ . Calculate the molar masses of butane ( $C_{4} H_{10}$ ) and oxygen ( $O_{2}$ ) and then determine the initial moles of each reactant.

M_{C_{4} H_{10}} = (4 \times 12) + (10 \times 1) = 58 g/mol

M_{O_{2}} = (2 \times 16) = 32 g/mol

n_{C_{4} H_{10}} = \frac{174.0 \times 10^{3} g}{58 g/mol} = 3000 mol

n_{O_{2}} = \frac{320.0 \times 10^{3} g}{32 g/mol} = 10000 mol

Step 2: Identify the limiting reagent and calculate moles of water formed.○ Expand

Determine the limiting reagent by comparing the mole ratio of reactants to their stoichiometric coefficients. Then, use the limiting reagent to calculate the moles of water ( $H_{2} O$ ) produced based on the balanced equation.

Ratio for C_{4} H_{10} = \frac{3000}{2} = 1500

Ratio for O_{2} = \frac{10000}{13} \approx 769.23

Since 769.23 < 1500, O_{2} is the limiting reagent.

From the balanced equation, 13 mol O_{2} produces 10 mol H_{2} O .

n_{H_{2} O} = 10000 mol O_{2} \times \frac{10 mol H_{2} O}{13 mol O_{2}} = \frac{100000}{13} mol

Step 3: Calculate the volume of water formed in liters.○ Expand

Calculate the mass of water formed using its molar mass. Then, convert the mass to volume using the given density of water ( $1 {g mL}^{- 1}$ ) and finally convert the volume to liters, rounding to the nearest integer.

M_{H_{2} O} = (2 \times 1) + 16 = 18 g/mol

Mass of H_{2} O = \frac{100000}{13} mol \times 18 g/mol = \frac{1800000}{13} g

Volume of H_{2} O in mL = \frac{1800000}{13} mL (since density is 1 g/mL)

Volume of H_{2} O in L = \frac{1800000}{13 \times 1000} L = \frac{1800}{13} L \approx 138.46 L

Rounding to the nearest integer, Volume of H_{2} O = 138 L

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Chemistry Question 71 – JEE-MAIN 2025

Clue 1 of 3: Stoichiometry and Limiting Reagent

🧠 Full Solution Path