1. Write the balanced reaction

For complete combustion of propane:

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

2. Calculate molar masses

• Propane, $\mathrm{C_3H_8}$:

$M(\mathrm{C_3H_8}) = 3(12.00) + 8(1.008) = 36.00 + 8.064 = 44.064\,\mathrm{g\,mol^{-1}}$

• Oxygen, $\mathrm{O_2}$:

$M(\mathrm{O_2}) = 2(16.00) = 32.00\,\mathrm{g\,mol^{-1}}$

3. Find initial moles

• Moles of propane:

$n(\mathrm{C_3H_8}) = \frac{100}{44.064} \approx 2.269\,\mathrm{mol}$

• Moles of oxygen:

$n(\mathrm{O_2}) = \frac{1000}{32.00} = 31.25\,\mathrm{mol}$

4. Determine the limiting reagent

From the balanced equation:

$1\,\text{mol } \mathrm{C_3H_8} \text{ requires } 5\,\text{mol } \mathrm{O_2}$

Oxygen needed for $2.269$ mol propane:

$5 \times 2.269 = 11.345\,\mathrm{mol}$

Available oxygen is $31.25$ mol, so oxygen is in excess and propane is the limiting reagent.

5. Calculate moles of products and excess reactant

From

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

For $2.269$ mol propane:

• Carbon dioxide formed:

$n(\mathrm{CO_2}) = 3 \times 2.269 = 6.807\,\mathrm{mol}$

• Water formed:

$n(\mathrm{H_2O}) = 4 \times 2.269 = 9.076\,\mathrm{mol}$

• Oxygen consumed:

$n(\mathrm{O_2\ consumed}) = 5 \times 2.269 = 11.345\,\mathrm{mol}$

• Oxygen left:

$n(\mathrm{O_2\ left}) = 31.25 - 11.345 = 19.905\,\mathrm{mol}$

6. Find total moles in the resulting mixture

The resulting mixture contains $\mathrm{CO_2}$, $\mathrm{H_2O}$, and excess $\mathrm{O_2}$.

$n_{\text{total}} = 6.807 + 9.076 + 19.905 = 35.788\,\mathrm{mol}$

7. Calculate mole fraction of carbon dioxide

$X_{\mathrm{CO_2}} = \frac{6.807}{35.788} \approx 0.1902$

Given that the mole fraction is written as:

$x \times 10^{-2}$

So,

$x \times 10^{-2} = 0.1902$

$x = 0.1902 \times 10^2 = 19.02$

Nearest integer:

$\boxed{19}$

8. Comparison with stored answer

Stored correct answer = $19$.

Our derived answer is also $19$, so it agrees.