Question

When 80 gm CH₄ is burnt, CO and CO₂ gases are formed in 1 : 4 mole ratio. If the mass of O₂ used in combustion is w gm then find value of (w/100).

Answer 1

3.04

Answer 2

The problem involves the incomplete combustion of methane (CH₄) leading to the formation of both carbon monoxide (CO) and carbon dioxide (CO₂). We need to determine the total mass of oxygen (O₂) consumed.

1. Write Balanced Chemical Equations: The combustion of methane can proceed via two pathways:

• To Carbon Monoxide (CO): $2CH_4(g) + 3O_2(g) \rightarrow 2CO(g) + 4H_2O(g)$ (From this, 1 mole of CH₄ reacts with 3/2 moles of O₂ to produce 1 mole of CO).

• To Carbon Dioxide (CO₂): $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$ (From this, 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂).

2. Calculate Moles of CH₄: Given mass of CH₄ = 80 gm. Molar mass of CH₄ = 12 (C) + 4 * 1 (H) = 16 g/mol. Moles of CH₄ = $\frac{\text{Mass}}{\text{Molar Mass}} = \frac{80 \text{ g}}{16 \text{ g/mol}} = 5 \text{ moles}$.

3. Determine Moles of CH₄ for each pathway: Let 'x' be the moles of CH₄ that react to form CO, and 'y' be the moles of CH₄ that react to form CO₂. Total moles of CH₄ = x + y = 5 moles.

According to the balanced equations:

• Moles of CO formed = x (since 1 mol CH₄ produces 1 mol CO in the first reaction stoichiometry when scaled down to 1 mol CH₄).
• Moles of CO₂ formed = y (since 1 mol CH₄ produces 1 mol CO₂ in the second reaction).

Given that the mole ratio of CO : CO₂ is 1 : 4. So, $\frac{\text{Moles of CO}}{\text{Moles of CO}_2} = \frac{x}{y} = \frac{1}{4}$. This implies y = 4x.

Substitute y = 4x into the total moles equation: x + 4x = 5 5x = 5 x = 1 mole.

Now find y: y = 4 * x = 4 * 1 = 4 moles.

So, 1 mole of CH₄ reacts to form CO, and 4 moles of CH₄ react to form CO₂.

4. Calculate Moles of O₂ consumed for each pathway:

• For CO formation (from 1 mole CH₄): From $2CH_4 + 3O_2 \rightarrow 2CO + 4H_2O$, for every 2 moles of CH₄, 3 moles of O₂ are consumed. Therefore, for 1 mole of CH₄, $\frac{3}{2} = 1.5$ moles of O₂ are consumed.

• For CO₂ formation (from 4 moles CH₄): From $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$, for every 1 mole of CH₄, 2 moles of O₂ are consumed. Therefore, for 4 moles of CH₄, $4 \times 2 = 8$ moles of O₂ are consumed.

5. Calculate Total Moles and Mass of O₂ consumed: Total moles of O₂ = (Moles of O₂ for CO) + (Moles of O₂ for CO₂) Total moles of O₂ = 1.5 moles + 8 moles = 9.5 moles.

Molar mass of O₂ = 2 * 16 = 32 g/mol. Mass of O₂ (w) = Total moles of O₂ * Molar mass of O₂ w = 9.5 moles * 32 g/mol = 304 g.

6. Calculate the value of (w/100): $\frac{w}{100} = \frac{304}{100} = 3.04$.

The final answer is 3.04

Explanation of the solution:

1. Balanced equations for CH₄ combustion to CO and CO₂ are established: $2CH_4 + 3O_2 \rightarrow 2CO + 4H_2O$ $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
2. Total moles of CH₄ are calculated: $80 \text{ g} / 16 \text{ g/mol} = 5 \text{ mol}$.
3. Let 'x' moles of CH₄ form CO and 'y' moles form CO₂. So, x + y = 5.
4. Given CO:CO₂ mole ratio is 1:4, implying x:y = 1:4, or y = 4x.
5. Solving x + 4x = 5 gives x = 1 mol and y = 4 mol.
6. Oxygen required for 1 mol CH₄ (to CO): $1 \times (3/2) = 1.5 \text{ mol } O_2$.
7. Oxygen required for 4 mol CH₄ (to CO₂): $4 \times 2 = 8 \text{ mol } O_2$.
8. Total O₂ moles = $1.5 + 8 = 9.5 \text{ mol}$.
9. Mass of O₂ (w) = $9.5 \text{ mol} \times 32 \text{ g/mol} = 304 \text{ g}$.
10. Value of (w/100) = $304/100 = 3.04$.

Answer:

The value of (w/100) is 3.04.