` 28.0L` of `CO_(2)` is produced on complete combustion of `16.8L` gaseous mixture of ethene and methane at `25^(@)C` and `1atm`.Heat evolved during the combustion process is
Given: `Delta H_(c)(CH_(4))=-900kJmol^(-1)` `Delta H_(c)(C_2H_(4)))=-1400kJmol^(-1)`

Answer

To solve the problem, we need to determine the heat evolved during the complete combustion of a gaseous mixture of ethene (C₂H₄) and methane (CH₄) that produces 28.0 L of CO₂ from a total of 16.8 L of the gaseous mixture. We will use the given enthalpy changes of combustion for both gases.

### Step-by-Step Solution:

1. **Identify the Combustion Reactions**:
- For methane (CH₄):
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
- For ethene (C₂H₄):
\[ \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \]

2. **Determine the Volume of Gases**:
- Let \( x \) be the volume of ethene (C₂H₄) in the mixture.
- Then, the volume of methane (CH₄) will be \( 16.8 - x \).
- The total volume of CO₂ produced from the combustion is given as 28.0 L.

3. **Calculate the Volume of CO₂ Produced**:
- From the combustion of ethene, the volume of CO₂ produced is \( 2x \).
- From the combustion of methane, the volume of CO₂ produced is \( 16.8 - x \).
- Therefore, the total volume of CO₂ produced can be expressed as:
\[ 2x + (16.8 - x) = 28.0 \]
- Simplifying this equation:
\[ x + 16.8 = 28.0 \]
\[ x = 28.0 - 16.8 = 11.2 \text{ L} \]

4. **Determine the Volume of Methane**:
- Now, substituting \( x \) back to find the volume of methane:
\[ \text{Volume of CH}_4 = 16.8 - x = 16.8 - 11.2 = 5.6 \text{ L} \]

5. **Calculate the Moles of Each Gas**:
- Using the ideal gas equation \( PV = nRT \), where \( P = 1 \text{ atm} \), \( R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1} \), and \( T = 298 \text{ K} \):
- Moles of CH₄:
\[ n_{\text{CH}_4} = \frac{PV}{RT} = \frac{1 \times 5.6}{0.082 \times 298} \approx 0.229 \text{ moles} \]
- Moles of C₂H₄:
\[ n_{\text{C}_2\text{H}_4} = \frac{PV}{RT} = \frac{1 \times 11.2}{0.082 \times 298} \approx 0.458 \text{ moles} \]

6. **Calculate the Heat Evolved**:
- Using the given enthalpy changes:
- For CH₄: \( \Delta H_c = -900 \text{ kJ/mol} \)
- For C₂H₄: \( \Delta H_c = -1400 \text{ kJ/mol} \)
- Total heat evolved:
\[ \text{Heat} = n_{\text{CH}_4} \times (-900) + n_{\text{C}_2\text{H}_4} \times (-1400) \]
\[ \text{Heat} = (0.229 \times -900) + (0.458 \times -1400) \]
\[ \text{Heat} = -206.1 - 641.2 = -847.3 \text{ kJ} \]

### Final Answer:
The heat evolved during the combustion process is **-847.3 kJ**.