A sample of coal gas contains 50% hydrogen, 30% methane, 14% CO and 6% `C_2H_4` 100 ml of this coal gas is mixed with 150 ml of oxygen and the mixture is exploded. What will be the composition of the resulting gas when it is to room temperature ?

Options

A. `52.81 %O_2 & 47.19% CO_2`
B. `47.33% O_2 & 52.67 % CO_2`
C. `41.67% O_2 & 58.33% CO_2`
D. `33.87% O_2 & 66.13 % CO_2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the volumes of each component in the coal gas.
The coal gas contains:
- 50% Hydrogen (H₂)
- 30% Methane (CH₄)
- 14% Carbon Monoxide (CO)
- 6% Ethylene (C₂H₄)

Given that the total volume of coal gas is 100 ml:
- Volume of H₂ = 50% of 100 ml = 50 ml
- Volume of CH₄ = 30% of 100 ml = 30 ml
- Volume of CO = 14% of 100 ml = 14 ml
- Volume of C₂H₄ = 6% of 100 ml = 6 ml

### Step 2: Write the balanced chemical reactions for the combustion of each component.
1. For Hydrogen:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]
(1 volume of O₂ is required for 2 volumes of H₂)

2. For Methane:
\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]
(2 volumes of O₂ are required for 1 volume of CH₄)

3. For Carbon Monoxide:
\[ 2CO + O_2 \rightarrow 2CO_2 \]
(1 volume of O₂ is required for 2 volumes of CO)

4. For Ethylene:
\[ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \]
(3 volumes of O₂ are required for 1 volume of C₂H₄)

### Step 3: Calculate the volume of O₂ required for the combustion of each component.
- For H₂ (50 ml):
\[ \text{O}_2 \text{ required} = \frac{1}{2} \times 50 \text{ ml} = 25 \text{ ml} \]

- For CH₄ (30 ml):
\[ \text{O}_2 \text{ required} = 2 \times 30 \text{ ml} = 60 \text{ ml} \]

- For CO (14 ml):
\[ \text{O}_2 \text{ required} = \frac{1}{2} \times 14 \text{ ml} = 7 \text{ ml} \]

- For C₂H₄ (6 ml):
\[ \text{O}_2 \text{ required} = 3 \times 6 \text{ ml} = 18 \text{ ml} \]

### Step 4: Calculate the total volume of O₂ required.
\[ \text{Total O}_2 \text{ required} = 25 \text{ ml} + 60 \text{ ml} + 7 \text{ ml} + 18 \text{ ml} = 110 \text{ ml} \]

### Step 5: Determine the unused volume of O₂.
Given that we started with 150 ml of O₂:
\[ \text{Unused O}_2 = 150 \text{ ml} - 110 \text{ ml} = 40 \text{ ml} \]

### Step 6: Calculate the volume of CO₂ produced.
- From CH₄: 30 ml of CH₄ produces 30 ml of CO₂.
- From CO: 14 ml of CO produces 14 ml of CO₂.
- From C₂H₄: 6 ml of C₂H₄ produces \(2 \times 6 = 12\) ml of CO₂.

Total volume of CO₂ produced:
\[ \text{Total CO}_2 = 30 \text{ ml} + 14 \text{ ml} + 12 \text{ ml} = 56 \text{ ml} \]

### Step 7: Determine the composition of the resulting gas.
The resulting gas consists of:
- CO₂: 56 ml
- Unused O₂: 40 ml

Total volume of resulting gas:
\[ \text{Total volume} = 56 \text{ ml} + 40 \text{ ml} = 96 \text{ ml} \]

### Step 8: Calculate the percentage composition of the resulting gas.
- Percentage of CO₂:
\[ \text{Percentage of CO}_2 = \left(\frac{56}{96}\right) \times 100 \approx 58.33\% \]

- Percentage of O₂:
\[ \text{Percentage of O}_2 = \left(\frac{40}{96}\right) \times 100 \approx 41.67\% \]

### Final Answer:
The composition of the resulting gas at room temperature is approximately:
- 58.33% CO₂
- 41.67% O₂

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