Let the mixture contain

• $x$ L of $\mathrm{CH_4}$
• $y$ L of $\mathrm{C_2H_4}$

Since total volume of the gaseous mixture is $16.8$ L, $x+y=16.8$

Use combustion stoichiometry

For methane: $\mathrm{CH_4 + 2O_2 \to CO_2 + 2H_2O}$ So, $1$ volume of $\mathrm{CH_4}$ gives $1$ volume of $\mathrm{CO_2}$.

For ethene: $\mathrm{C_2H_4 + 3O_2 \to 2CO_2 + 2H_2O}$ So, $1$ volume of $\mathrm{C_2H_4}$ gives $2$ volumes of $\mathrm{CO_2}$.

Total $\mathrm{CO_2}$ produced is $28.0$ L, so $x+2y=28.0$

Solve the two equations

From $x+y=16.8$ $x+2y=28.0$

Subtracting, $y=28.0-16.8=11.2$ Hence, $x=16.8-11.2=5.6$

Therefore, the mixture contains:

• $5.6$ L $\mathrm{CH_4}$
• $11.2$ L $\mathrm{C_2H_4}$

Convert volumes to moles

At $25^\circ$C and $1$ atm, molar volume is approximately $24.0\ \text{L mol}^{-1}$

So, $n(\mathrm{CH_4})=\frac{5.6}{24}=0.2333\ \text{mol}$ $n(\mathrm{C_2H_4})=\frac{11.2}{24}=0.4667\ \text{mol}$

Calculate heat evolved

Given: $\Delta H_c(\mathrm{CH_4})=-900\ \text{kJ mol}^{-1}$ $\Delta H_c(\mathrm{C_2H_4})=-1400\ \text{kJ mol}^{-1}$

Total enthalpy change: $\Delta H=0.2333(-900)+0.4667(-1400)$

$\Delta H=-210-653.38$ $\Delta H\approx -863.38\ \text{kJ}$

Since the question asks for heat evolved, we report the magnitude: $\boxed{863\ \text{kJ}}$

Comparison with stored answer

Stored correct answer = $847$

My derived answer is $863$, which does not match $847$.

The standard approach uses the molar volume at $25^\circ$C and $1$ atm, i.e. about $24.0$ L mol$^{-1}$, giving $863$ kJ. Hence the stored answer likely comes from using an incorrect molar volume or rounding inconsistency.