When 5 litres of a gas mixture of methane and propane is perfectly combusted at `0^(@)C` and 1 atmosphere, 16 litre of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ`(Delta H_(comb) (CH_(4)) = 890 kJ mol^(-1), Delta H_(comb) (C_(3)H_(8)) = 2220 kJ mol^(-1))` is

Options

A. 32
B. 38
C. 317
D. 477

Answer

To solve the problem of calculating the heat released from the combustion of a gas mixture of methane and propane, we can follow these steps:

### Step 1: Write the Combustion Reactions
The combustion reactions for methane (CH₄) and propane (C₃H₈) are as follows:

1. **Methane:**
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \quad \Delta H_{\text{comb}} = -890 \text{ kJ/mol} \]

2. **Propane:**
\[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \quad \Delta H_{\text{comb}} = -2220 \text{ kJ/mol} \]

### Step 2: Determine the Volume of Each Gas
Let \( x \) be the volume of methane (in liters) and \( 5 - x \) be the volume of propane. Given that 16 liters of oxygen are consumed, we can set up the stoichiometric relationships based on the reactions.

From the combustion reactions:
- Methane consumes 2 liters of O₂ for every 1 liter of CH₄.
- Propane consumes 5 liters of O₂ for every 1 liter of C₃H₈.

The equation for the total oxygen consumed is:
\[ 2x + 5(5 - x) = 16 \]

### Step 3: Solve for \( x \)
Expanding the equation:
\[ 2x + 25 - 5x = 16 \]
\[ -3x + 25 = 16 \]
\[ -3x = 16 - 25 \]
\[ -3x = -9 \]
\[ x = 3 \]

Thus, the volume of methane is \( 3 \) liters, and the volume of propane is:
\[ 5 - 3 = 2 \text{ liters} \]

### Step 4: Calculate Moles of Each Gas
Using the ideal gas law, at standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, we can calculate the moles of each gas:

- Moles of methane:
\[ \text{Moles of CH}_4 = \frac{3 \text{ L}}{22.4 \text{ L/mol}} = 0.134 \text{ mol} \]

- Moles of propane:
\[ \text{Moles of C}_3\text{H}_8 = \frac{2 \text{ L}}{22.4 \text{ L/mol}} = 0.0893 \text{ mol} \]

### Step 5: Calculate Heat Released
Now we can calculate the total heat released from the combustion of both gases using their respective heats of combustion:

Total heat released (\( Q \)):
\[ Q = (\text{Moles of CH}_4 \times \Delta H_{\text{comb}}(\text{CH}_4)) + (\text{Moles of C}_3\text{H}_8 \times \Delta H_{\text{comb}}(\text{C}_3\text{H}_8)) \]
Substituting the values:
\[ Q = (0.134 \text{ mol} \times -890 \text{ kJ/mol}) + (0.0893 \text{ mol} \times -2220 \text{ kJ/mol}) \]
Calculating each term:
\[ Q = -119.26 \text{ kJ} + -198.86 \text{ kJ} \]
\[ Q = -318.12 \text{ kJ} \]

### Final Answer
The total heat released from the combustion of the gas mixture is approximately:
\[ Q \approx 318.12 \text{ kJ} \]