Logo
Jeeify
NVAMediumJEE 2025Stoichiometry & Calculations

JEE Chemistry 2025 Question with Solution

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

C4H10(g)+132O2(g)→4CO2(g)+5H2O(l)C_4H_{10}(g) + \frac{13}{2} O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l)

If 174.0 kg174.0 \, \text{kg} of butane is mixed with 320.0 kg320.0 \, \text{kg} of O2O_2, the volume of water formed in litres is _____. (Nearest integer)

[Given: (a) Molar masses: C = 1212, H = 11, O = 16 g mol−116 \, \text{g} \, \text{mol}^{-1}; (b) Density of water = 1 g mL−11 \, \text{g} \, \text{mL}^{-1}]

Answer

Correct answer:138

Step-by-step solution

Standard Method

Given:

  • Butane combustion reaction:
C4H10(g)+132O2(g)→4CO2(g)+5H2O(l)C_4H_{10}(g) + \frac{13}{2} O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l)
  • Mass of butane = 174×103 g174 \times 10^3 \, \text{g}
  • Mass of oxygen = 320×103 g320 \times 10^3 \, \text{g}
  • Density of water = 1 g mL−11 \, \text{g} \, \text{mL}^{-1}

Find: The volume of water formed in litres.

Step 1: Calculate moles of reactants.

nC4H10=174×10358=3000 moln_{C_4H_{10}} = \frac{174 \times 10^3}{58} = 3000 \, \text{mol} nO2=320×10332=10000 moln_{O_2} = \frac{320 \times 10^3}{32} = 10000 \, \text{mol}

Step 2: Identify the limiting reactant. From the balanced equation, 11 mol C4H10C_4H_{10} reacts with 6.56.5 mol O2O_2.

3000×6.5=19500 mol O23000 \times 6.5 = 19500 \, \text{mol } O_2

Since only 10000 mol10000 \, \text{mol} of O2O_2 is available, oxygen is the limiting reactant.

Step 3: Calculate moles of water formed. From stoichiometry, 6.56.5 mol O2O_2 produces 55 mol H2OH_2O.

nH2O=56.5×10000=1013×10000≈7692.3 moln_{H_2O} = \frac{5}{6.5} \times 10000 = \frac{10}{13} \times 10000 \approx 7692.3 \, \text{mol}

Step 4: Calculate the volume of water formed.

mH2O=7692.3×18≈138461.4 gm_{H_2O} = 7692.3 \times 18 \approx 138461.4 \, \text{g}

Using density of water,

VH2O=138461.4 g1 g mL−1=138461.4 mL=138.4614 LV_{H_2O} = \frac{138461.4 \, \text{g}}{1 \, \text{g} \, \text{mL}^{-1}} = 138461.4 \, \text{mL} = 138.4614 \, \text{L}

Nearest integer = 138138.

Therefore, the volume of water formed is 138 L138 \, \text{L}.

Limiting Reactant Route

Given: Masses of butane and oxygen are provided, and water is to be found using stoichiometry and density.

Find: Volume of water formed.

First compute the reactant moles:

nC4H10=17400058=3000  moln_{C_4H_{10}} = \frac{174000}{58} = 3000 \; \text{mol} nO2=32000032=10000  moln_{O_2} = \frac{320000}{32} = 10000 \; \text{mol}

For 30003000 mol butane, required oxygen is:

3000×6.5=19500  mol3000 \times 6.5 = 19500 \; \text{mol}

This is greater than the available 1000010000 mol, so O2O_2 is limiting.

Moles of butane that can actually react are:

nreact=100006.5=1538.46  moln_{\text{react}} = \frac{10000}{6.5} = 1538.46 \; \text{mol}

From the equation, 11 mol butane gives 55 mol water:

nH2O=5×1538.46=7692.31  moln_{H_2O} = 5 \times 1538.46 = 7692.31 \; \text{mol}

Mass of water formed:

mH2O=7692.31×18=138461.6  gm_{H_2O} = 7692.31 \times 18 = 138461.6 \; \text{g}

Since density of water is 1 g mL−11 \, \text{g} \, \text{mL}^{-1}, the numerical value of mass in grams equals the volume in millilitres:

V=138461.6  mL=138.46  LV = 138461.6 \; \text{mL} = 138.46 \; \text{L}

Therefore, the nearest integer value is 138138.

Common mistakes

  • A common mistake is treating butane as the limiting reactant without comparing mole ratios. This is wrong because the balanced equation requires 6.56.5 mol of O2O_2 per mol of C4H10C_4H_{10}. Always calculate the required amount of one reactant for the available amount of the other before deciding the limiting reactant.

  • Another mistake is using the given masses directly in the stoichiometric ratio. This is incorrect because stoichiometric coefficients apply to moles, not masses. First convert every given mass into moles using molar mass.

  • Students may convert water mass incorrectly into volume by forgetting the density relation or by mixing mL and L. Since density is 1 g mL−11 \, \text{g} \, \text{mL}^{-1}, the mass in grams equals the volume in millilitres. Then divide by 10001000 to convert millilitres into litres.

Practice more Stoichiometry & Calculations questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Stoichiometry & Calculations

Related questions