100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is $$x \times 10^{-2}$$. The value of $$x$$ is _________. (Nearest integer)

We have been supplied with 100 g of propane, whose molecular formula is $$\mathrm{C_3H_8}$$. The molar mass of propane is obtained by adding the atomic masses: $$3(12\ \text{g mol}^{-1}) + 8(1\ \text{g mol}^{-1}) = 44\ \text{g mol}^{-1}.$$

The general formula for finding moles is first stated:

$$\text{Number of moles} = \dfrac{\text{Given mass}}{\text{Molar mass}}.$$

Using this, the number of moles of propane present is

$$n_{\mathrm{C_3H_8}} = \dfrac{100\ \text{g}}{44\ \text{g mol}^{-1}} = 2.2727\ \text{mol}.$$

Next, the oxygen. The molecular mass of $$\mathrm{O_2}$$ is $$32\ \text{g mol}^{-1}.$$ Hence,

$$n_{\mathrm{O_2,\ given}} = \dfrac{1000\ \text{g}}{32\ \text{g mol}^{-1}} = 31.25\ \text{mol}.$$

The balanced combustion equation for propane is written and will now be used for stoichiometry:

$$\mathrm{C_3H_8 + 5\,O_2 \;\rightarrow\; 3\,CO_2 + 4\,H_2O}.$$

From the coefficients we see that 1 mol of propane needs 5 mol of oxygen. Therefore, the oxygen required for the full consumption of the available propane is

$$n_{\mathrm{O_2,\ reqd}} = 5 \times n_{\mathrm{C_3H_8}} = 5 \times 2.2727 = 11.3636\ \text{mol}.$$

Because $$31.25\ \text{mol} > 11.3636\ \text{mol},$$ oxygen is in excess and propane is the limiting reagent. Hence all propane will be consumed.

By direct proportion from the balanced equation, the amounts of products formed are:

• Carbon dioxide: $$n_{\mathrm{CO_2}} = 3 \times n_{\mathrm{C_3H_8}} = 3 \times 2.2727 = 6.8182\ \text{mol}.$$
• Water vapour: $$n_{\mathrm{H_2O}} = 4 \times n_{\mathrm{C_3H_8}} = 4 \times 2.2727 = 9.0909\ \text{mol}.$$

The oxygen actually consumed equals $$11.3636\ \text{mol},$$ so the oxygen remaining after reaction is

$$n_{\mathrm{O_2,\ left}} = 31.25 - 11.3636 = 19.8864\ \text{mol}.$$

The gaseous mixture after completion therefore contains:

$$\begin{aligned} n_{\mathrm{CO_2}} &= 6.8182\ \text{mol},\\ n_{\mathrm{H_2O}} &= 9.0909\ \text{mol},\\ n_{\mathrm{O_2}} &= 19.8864\ \text{mol}. \end{aligned}$$

The total number of moles in the mixture is obtained by simple addition:

$$n_{\text{total}} = 6.8182 + 9.0909 + 19.8864 = 35.7955\ \text{mol}.$$

We now recall the definition of mole fraction:

$$\chi_i = \dfrac{n_i}{n_{\text{total}}}.$$

Applying this to carbon dioxide,

$$\chi_{\mathrm{CO_2}} = \dfrac{6.8182}{35.7955} = 0.1904.$$

The question states that this value is written as $$x \times 10^{-2},$$ so

$$x = 0.1904 \times 10^{2} = 19.04.$$

Rounding to the nearest integer gives $$x = 19.$$

So, the answer is $$19$$.