Questions: Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 2.74 g of water is produced from the reaction of 10.2 g of ethane and 18.3 g of oxygen gas, calculate the percent yield of water. Round your answer to 3 significant figures.

Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 2.74 g of water is produced from the reaction of 10.2 g of ethane and 18.3 g of oxygen gas, calculate the percent yield of water.

Round your answer to 3 significant figures.

Transcript text: Gaseous ethane $\left(\mathrm{CH}_{3} \mathrm{CH}_{3}\right)$ reacts with gaseous oxygen gas $\left(\mathrm{O}_{2}\right)$ to produce gaseous carbon dioxide $\left(\mathrm{CO}_{2}\right)$ and gaseous water $\left(\mathrm{H}_{2} \mathrm{O}\right)$. If 2.74 g of water is produced from the reaction of 10.2 g of ethane and 18.3 g of oxygen gas, calculate the percent yield of water. Round your answer to 3 significant figures.

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the combustion of ethane is: \[ 2 \mathrm{C}_2\mathrm{H}_6 + 7 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + 6 \mathrm{H}_2\mathrm{O} \]

Step 2: Calculate Moles of Reactants

First, we need to calculate the moles of ethane and oxygen gas.

Molar mass of ethane ($\mathrm{C}_2\mathrm{H}_6$): \[ 2 \times 12.01 + 6 \times 1.008 = 30.07 \, \text{g/mol} \] \[ \text{Moles of ethane} = \frac{10.2 \, \text{g}}{30.07 \, \text{g/mol}} = 0.3392 \, \text{mol} \]
Molar mass of oxygen gas ($\mathrm{O}_2$): \[ 2 \times 16.00 = 32.00 \, \text{g/mol} \] \[ \text{Moles of oxygen gas} = \frac{18.3 \, \text{g}}{32.00 \, \text{g/mol}} = 0.5719 \, \text{mol} \]

Step 3: Determine the Limiting Reactant

Using the stoichiometry of the balanced equation, we find the limiting reactant.

According to the balanced equation, 2 moles of $\mathrm{C}_2\mathrm{H}_6$ react with 7 moles of $\mathrm{O}_2$. \[ \text{Required moles of } \mathrm{O}_2 = 0.3392 \, \text{mol} \times \frac{7}{2} = 1.1872 \, \text{mol} \]

Since we only have 0.5719 moles of $\mathrm{O}_2$, oxygen gas is the limiting reactant.

Step 4: Calculate Theoretical Yield of Water

Using the limiting reactant, calculate the theoretical yield of water.

According to the balanced equation, 7 moles of $\mathrm{O}_2$ produce 6 moles of $\mathrm{H}_2\mathrm{O}$. \[ \text{Moles of } \mathrm{H}_2\mathrm{O} = 0.5719 \, \text{mol} \times \frac{6}{7} = 0.4902 \, \text{mol} \]
Molar mass of water ($\mathrm{H}_2\mathrm{O}$): \[ 2 \times 1.008 + 16.00 = 18.02 \, \text{g/mol} \] \[ \text{Theoretical yield of water} = 0.4902 \, \text{mol} \times 18.02 \, \text{g/mol} = 8.8316 \, \text{g} \]

Step 5: Calculate Percent Yield

Finally, calculate the percent yield of water. \[ \text{Percent yield} = \left( \frac{2.74 \, \text{g}}{8.8316 \, \text{g}} \right) \times 100\% = 31.0\% \]

Final Answer

\[ \boxed{31.0\%} \]

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