28.0 L of CO$$_2$$ is produced on complete combustion of 16.8 L gaseous mixture of ethene and methane at 25°C and 1 atm. Heat evolved during the combustion process is _____ kJ
Given: $$\Delta H_C$$(CH$$_4$$) = -900 kJ mol$$^{-1}$$
$$\Delta H_C$$(C$$_2$$H$$_4$$) = -1400 kJ mol$$^{-1}$$.

By Avogadro's law, for gases at the same temperature and pressure, the volume ratio is equal to the mole ratio.

The balanced combustion reactions are:

$$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$$

$$C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)$$

Let the volumes of methane and ethene in the mixture be $$x$$ L and $$y$$ L, respectively.

Since the total volume of the mixture is $$16.8$$ L,

$$x+y=16.8$$

From the stoichiometry of the reactions, $$1$$ volume of $$CH_4$$ produces $$1$$ volume of $$CO_2$$, whereas $$1$$ volume of $$C_2H_4$$ produces $$2$$ volumes of $$CO_2$$.

Hence,

$$x+2y=28.0$$

Subtracting the first equation from the second,

$$(x+2y)-(x+y)=28.0-16.8$$

$$y=11.2\ \text{L}$$

Substituting into $$x+y=16.8$$,

$$x+11.2=16.8$$

$$x=5.6\ \text{L}$$

Therefore,

• Volume of $$CH_4 = 5.6$$ L
• Volume of $$C_2H_4 = 11.2$$ L

At $$25^\circ\text{C}$$ and $$1$$ atm,

$$V_m=\frac{RT}{P}=\frac{0.0821\times298}{1}\approx24.46\ \text{L mol}^{-1}$$

The number of moles of each gas is

$$n(CH_4)=\frac{5.6}{24.46}\approx0.229\ \text{mol}$$

$$n(C_2H_4)=\frac{11.2}{24.46}\approx0.458\ \text{mol}$$

The total heat evolved is

$$\text{Heat}=\left(0.229\times(-900)\right)+\left(0.458\times(-1400)\right)$$

$$=-206.1-641.2$$

$$=-847.3\ \text{kJ}$$

Hence, the magnitude of heat evolved is

$$\boxed{847.3\ \text{kJ}}$$

If the simplified molar volume of $$22.4\ \text{L mol}^{-1}$$ (STP) is used instead, the heat evolved is approximately $$925\ \text{kJ}$$.