Questions: Liquid octane (CH3(CH2)6CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 1.26 g of water is produced from the reaction of 2.28 g of octane and 15.5 g of oxygen gas, calculate the percent yield of water. Round your answer to 3 significant figures.

Solution Steps

The balanced chemical equation for the combustion of octane is:

\[ 2 \, \text{C}_8\text{H}_{18} + 25 \, \text{O}_2 \rightarrow 16 \, \text{CO}_2 + 18 \, \text{H}_2\text{O} \]

• Molar mass of octane (\(\text{C}_8\text{H}_{18}\)): \[ (8 \times 12.01) + (18 \times 1.008) = 114.22 \, \text{g/mol} \]

• Molar mass of water (\(\text{H}_2\text{O}\)): \[ (2 \times 1.008) + 16.00 = 18.016 \, \text{g/mol} \]

• Moles of octane: \[ \frac{2.28 \, \text{g}}{114.22 \, \text{g/mol}} = 0.01996 \, \text{mol} \]

• Moles of oxygen: \[ \frac{15.5 \, \text{g}}{32.00 \, \text{g/mol}} = 0.4844 \, \text{mol} \]

From the balanced equation, 2 moles of octane react with 25 moles of oxygen. Therefore, the ratio is:

\[ \frac{0.01996 \, \text{mol C}_8\text{H}_{18}}{2} = 0.00998 \]

\[ \frac{0.4844 \, \text{mol O}_2}{25} = 0.01938 \]

Since \(0.00998 < 0.01938\), octane is the limiting reactant.

From the balanced equation, 2 moles of octane produce 18 moles of water. Therefore, the moles of water produced are:

\[ 0.01996 \, \text{mol C}_8\text{H}_{18} \times \frac{18 \, \text{mol H}_2\text{O}}{2 \, \text{mol C}_8\text{H}_{18}} = 0.1796 \, \text{mol H}_2\text{O} \]

The theoretical mass of water is:

\[ 0.1796 \, \text{mol} \times 18.016 \, \text{g/mol} = 3.236 \, \text{g} \]

The percent yield is given by:

\[ \text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\% \]

\[ \text{Percent Yield} = \left(\frac{1.26 \, \text{g}}{3.236 \, \text{g}}\right) \times 100\% = 38.94\% \]

Final Answer