Questions: If 2.00 grams of CO2 were produced when 4.00 grams of C5H12 were combusted in excess oxygen, what is the percent yield of the combustion reaction? A) 12.2 % B) 16.4 % C) 22.4 % D) 48.8 % E) 82.0 %

If 2.00 grams of CO2 were produced when 4.00 grams of C5H12 were combusted in excess oxygen, what is the percent yield of the combustion reaction?
A) 12.2 %
B) 16.4 %
C) 22.4 %
D) 48.8 %
E) 82.0 %

Transcript text: 3. If 2.00 grams of $\mathrm{CO}_{2}$ were produced when 4.00 grams of $\mathrm{C}_{5} \mathrm{H}_{12}$ were combusted in excess oxygen, what is the percent yield of the combustion reaction? A) $12.2 \%$ B) $16.4 \%$ C) $22.4 \%$ D) $48.8 \%$ E) $82.0 \%$

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the combustion of pentane ($\mathrm{C}_5\mathrm{H}_{12}$) is: \[ \mathrm{C}_5\mathrm{H}_{12} + 8 \mathrm{O}_2 \rightarrow 5 \mathrm{CO}_2 + 6 \mathrm{H}_2\mathrm{O} \]

Step 2: Calculate the Molar Masses

Molar mass of $\mathrm{C}_5\mathrm{H}_{12}$: \[ 5 \times 12.01 + 12 \times 1.008 = 60.05 + 12.096 = 72.15 \, \text{g/mol} \]
Molar mass of $\mathrm{CO}_2$: \[ 12.01 + 2 \times 16.00 = 12.01 + 32.00 = 44.01 \, \text{g/mol} \]

Step 3: Calculate the Theoretical Yield of $\mathrm{CO}_2$

Moles of $\mathrm{C}_5\mathrm{H}_{12}$ combusted: \[ \frac{4.00 \, \text{g}}{72.15 \, \text{g/mol}} = 0.0554 \, \text{mol} \]
According to the balanced equation, 1 mole of $\mathrm{C}_5\mathrm{H}_{12}$ produces 5 moles of $\mathrm{CO}_2$. Therefore, moles of $\mathrm{CO}_2$ produced: \[ 0.0554 \, \text{mol} \times 5 = 0.2770 \, \text{mol} \]
Theoretical mass of $\mathrm{CO}_2$: \[ 0.2770 \, \text{mol} \times 44.01 \, \text{g/mol} = 12.19 \, \text{g} \]

Step 4: Calculate the Percent Yield