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 step by step, we will follow the outlined process from the video transcript:

### Step 1: Write the Balanced Equations for Combustion
1. **For Methane (CH₄)**:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
The enthalpy change for combustion of methane is given as \(\Delta H_{\text{comb}}(\text{CH}_4) = -890 \, \text{kJ/mol}\).

2. **For Propane (C₃H₈)**:
\[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]
The enthalpy change for combustion of propane is given as \(\Delta H_{\text{comb}}(\text{C}_3\text{H}_8) = -2220 \, \text{kJ/mol}\).

### Step 2: Define Variables for the Gas Mixture
Let:
- \(x\) = volume of methane (CH₄) in liters.
- \(5 - x\) = volume of propane (C₃H₈) in liters.

### Step 3: Use Stoichiometry to Relate Volumes
From the problem, we know that 16 liters of oxygen is consumed. The stoichiometry of the reactions gives us the following equation:
\[ 2x + 5(5 - x) = 16 \]

### Step 4: Solve for \(x\)
Expanding the equation:
\[ 2x + 25 - 5x = 16 \]
Combining like terms:
\[ 25 - 3x = 16 \]
Solving for \(x\):
\[ 3x = 25 - 16 \implies 3x = 9 \implies x = 3 \, \text{liters} \]

### Step 5: Determine Volumes of Methane and Propane
- Volume of methane (CH₄) = \(x = 3\) liters.
- Volume of propane (C₃H₈) = \(5 - x = 5 - 3 = 2\) liters.

### Step 6: Calculate Moles of Each Gas
Using the molar volume at STP (22.4 L/mol):
- Moles of methane (CH₄):
\[ \text{Moles of CH}_4 = \frac{3 \, \text{liters}}{22.4 \, \text{liters/mol}} = \frac{3}{22.4} \, \text{mol} \]
- Moles of propane (C₃H₈):
\[ \text{Moles of C}_3\text{H}_8 = \frac{2 \, \text{liters}}{22.4 \, \text{liters/mol}} = \frac{2}{22.4} \, \text{mol} \]

### Step 7: Calculate Heat Released from Combustion
Using the enthalpy values:
\[ \text{Heat released} = \left(\frac{3}{22.4} \times 890\right) + \left(\frac{2}{22.4} \times 2220\right) \]
Calculating each term:
1. For methane:
\[ \frac{3}{22.4} \times 890 = \frac{2670}{22.4} \approx 119.64 \, \text{kJ} \]
2. For propane:
\[ \frac{2}{22.4} \times 2220 = \frac{4440}{22.4} \approx 198.21 \, \text{kJ} \]

### Step 8: Total Heat Released
Adding the two amounts:
\[ \text{Total Heat} = 119.64 + 198.21 \approx 317.85 \, \text{kJ} \]

### Conclusion
The amount of heat released from this combustion is approximately \(317 \, \text{kJ}\).