JEE Challenger
More from Some Basic Concepts of Chemistry
ChemMedMain 2026 • 2nd Apr - Shift 2

Mass of Carbon Dioxide Formed from Hydrocarbon Combustion

The ratio of mass percentage (w/w) of C:H\text{C} : \text{H} in a hydrocarbon is 12:112 : 1. It has two carbon atoms. The weight (in g) of CO2(g)\text{CO}_2(\text{g}) formed when 3.38 g3.38\text{ g} of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in g mol−1\text{g mol}^{-1} C:12,H:1,O:16\text{C} : 12, \text{H} : 1, \text{O} : 16)

Options

A

5.68

B

11.44

Correct
C

22.74

D

17.05

Topics & Concepts

Some Basic Concepts of ChemistryMole ConceptStoichiometry

Step-by-Step Solution

To find the weight of CO2\text{CO}_2 formed, we first determine the molecular formula of the hydrocarbon.

Step 1: Determine the Empirical Formula

The ratio of mass percentages (w/w) of C:H\text{C} : \text{H} in the hydrocarbon is given as 12:112 : 1.

The ratio of the number of moles of carbon and hydrogen atoms is: Moles of C:Moles of H=1212:11=1:1\text{Moles of C} : \text{Moles of H} = \frac{12}{12} : \frac{1}{1} = 1 : 1

Thus, the empirical formula of the hydrocarbon is CH\text{CH}.

Step 2: Determine the Molecular Formula

It is given that the hydrocarbon contains two carbon atoms. Therefore, the molecular formula of the hydrocarbon is C2H2\text{C}_2\text{H}_2 (ethyne).

Step 3: Calculate the Molar Mass of the Hydrocarbon

Molar mass of C2H2=(2×12)+(2×1)=26 g mol−1\text{Molar mass of }\text{C}_2\text{H}_2 = (2 \times 12) + (2 \times 1) = 26 \text{ g mol}^{-1}

Step 4: Calculate Moles of Hydrocarbon Used

The mass of C2H2\text{C}_2\text{H}_2 given is 3.38 g3.38\text{ g}. Moles of C2H2=3.38 g26 g mol−1=0.13 mol\text{Moles of }\text{C}_2\text{H}_2 = \frac{3.38\text{ g}}{26\text{ g mol}^{-1}} = 0.13\text{ mol}

Step 5: Calculate Moles and Mass of CO2\text{CO}_2 Formed

The complete combustion equation for C2H2\text{C}_2\text{H}_2 is: C2H2(g)+52O2(g)⟶2CO2(g)+H2O(l)\text{C}_2\text{H}_2(\text{g}) + \frac{5}{2}\text{O}_2(\text{g}) \longrightarrow 2\text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

From stoichiometry, 1 mole1\text{ mole} of C2H2\text{C}_2\text{H}_2 produces 2 moles2\text{ moles} of CO2\text{CO}_2. Moles of CO2 formed=2×0.13 mol=0.26 mol\text{Moles of }\text{CO}_2\text{ formed} = 2 \times 0.13\text{ mol} = 0.26\text{ mol}

The molar mass of CO2\text{CO}_2 is: Molar mass of CO2=12+(2×16)=44 g mol−1\text{Molar mass of }\text{CO}_2 = 12 + (2 \times 16) = 44\text{ g mol}^{-1}

Thus, the weight of CO2\text{CO}_2 formed is: Weight of CO2=0.26 mol×44 g mol−1=11.44 g\text{Weight of }\text{CO}_2 = 0.26\text{ mol} \times 44\text{ g mol}^{-1} = 11.44\text{ g}

Correct Option: B

Mass of Carbon Dioxide Formed from Hydrocarbon Combustion | Chemistry PYQ Solution - JEE Challenger