Question

Question was provided via image upload.

Explanation:

Step1: Balance the combustion equation

Balance C, H, then O:

$$2\text{C}_8\text{H}_{18} + 25\text{O}_2 ightarrow 16\text{CO}_2 + 18\text{H}_2\text{O}$$

Step2: Calculate molar masses

Molar masses:
$M(\text{C}_8\text{H}_{18}) = 8\times12.01 + 18\times1.008 = 114.23\ \text{g/mol}$
$M(\text{O}_2) = 2\times16.00 = 32.00\ \text{g/mol}$
$M(\text{CO}_2) = 12.01 + 2\times16.00 = 44.01\ \text{g/mol}$

Step3: Find moles of reactants

Moles of octane:
$$n(\text{C}_8\text{H}_{18}) = \frac{25.0\ \text{g}}{114.23\ \text{g/mol}} = 0.2189\ \text{mol}$$
Moles of oxygen:
$$n(\text{O}_2) = \frac{35.7\ \text{g}}{32.00\ \text{g/mol}} = 1.1156\ \text{mol}$$

Step4: Identify limiting reactant

From balanced equation, mole ratio $\frac{n(\text{O}_2)}{n(\text{C}_8\text{H}_{18})} = \frac{25}{2} = 12.5$
Required $\text{O}_2$ for given octane:
$$n(\text{O}_2)_{\text{req}} = 0.2189\ \text{mol} \times 12.5 = 2.736\ \text{mol}$$
Available $\text{O}_2$ (1.1156 mol) < required, so $\text{O}_2$ is limiting.

Step5: Calculate theoretical yield of $\text{CO}_2$

Mole ratio $\frac{n(\text{CO}_2)}{n(\text{O}_2)} = \frac{16}{25}$
$$n(\text{CO}_2)_{\text{theo}} = 1.1156\ \text{mol} \times \frac{16}{25} = 0.71398\ \text{mol}$$
$$m(\text{CO}_2)_{\text{theo}} = 0.71398\ \text{mol} \times 44.01\ \text{g/mol} = 31.42\ \text{g}$$

Step6: Calculate percent yield

$$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% = \frac{29.5\ \text{g}}{31.42\ \text{g}} \times 100\% = 93.9\%$$

Step7: Calculate percent error

$$\text{Percent Error} = \frac{|\text{Actual - Theoretical}|}{\text{Theoretical}} \times 100\% = \frac{|29.5 - 31.42|}{31.42} \times 100\% = 6.11\%$$

Step8: Find excess reactant remaining

Moles of octane consumed:
$$n(\text{C}_8\text{H}_{18})_{\text{used}} = 1.1156\ \text{mol} \times \frac{2}{25} = 0.08925\ \text{mol}$$
Moles of octane left:
$$n(\text{C}_8\text{H}_{18})_{\text{left}} = 0.2189 - 0.08925 = 0.12965\ \text{mol}$$
Mass of octane left:
$$m(\text{C}_8\text{H}_{18})_{\text{left}} = 0.12965\ \text{mol} \times 114.23\ \text{g/mol} = 14.81\ \text{g}$$

Answer:

Limiting Reactant: $\text{O}_2$ (Oxygen gas)
Theoretical Yield: $31.4\ \text{g}$
Percent Yield: $93.9\%$
Percent Error: $6.11\%$
Grams of excess reactant left over: $14.8\ \text{g}$