Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). if 18.2 g of water is produced from the reaction of 13.5 g of methane and 86.6 g of oxygen gas, calculate the percent yield of water.

Question

Be sure your answer has the correct number of significant digits in it.

Answer 1

To calculate the percent yield, we first need to determine the theoretical yield of water based on the amount of reactants used and then compare it to the actual yield obtained from the reaction.

1. **Write the balanced chemical equation:**
\[ \text{CH}_4 (g) + 2\text{O}_2 (g) \rightarrow \text{CO}_2 (g) + 2\text{H}_2\text{O} (g) \]

2. **Calculate the molar masses:**
- \( \text{CH}_4 \): \(12.01 \, (\text{C}) + 4 \times 1.01 \, (\text{H}) = 16.05 \, \text{g/mol} \)
- \( \text{O}_2 \): \(2 \times 16.00 \, (\text{O}) = 32.00 \, \text{g/mol} \)
- \( \text{H}_2\text{O} \): \(2 \times 1.01 \, (\text{H}) + 16.00 \, (\text{O}) = 18.02 \, \text{g/mol} \)

3. **Convert masses to moles:**
- Moles of \( \text{CH}_4 \): \(\frac{13.5 \, \text{g}}{16.05 \, \text{g/mol}} \approx 0.841 \, \text{mol} \)
- Moles of \( \text{O}_2 \): \(\frac{86.6 \, \text{g}}{32.00 \, \text{g/mol}} \approx 2.71 \, \text{mol} \)

4. **Determine the limiting reactant:**
- According to the balanced equation, 1 mole of \( \text{CH}_4 \) reacts with 2 moles of \( \text{O}_2 \).
- For 0.841 moles of \( \text{CH}_4 \), the required moles of \( \text{O}_2 \) is \(0.841 \times 2 = 1.682 \, \text{mol} \).
- We have 2.71 moles of \( \text{O}_2 \), which is more than sufficient to react with 0.841 moles of \( \text{CH}_4 \).

Therefore, \( \text{CH}_4 \) is the limiting reactant.

5. **Calculate theoretical yield:**
- According to the balanced equation, 1 mole of \( \text{CH}_4 \) produces 2 moles of \( \text{H}_2\text{O} \).
- Therefore, \( 0.841 \, \text{mol} \, \text{CH}_4 \) will produce \( 0.841 \times 2 = 1.682 \, \text{mol} \, \text{H}_2\text{O} \).
- Now, convert moles of \( \text{H}_2\text{O} \) to grams: \( 1.682 \, \text{mol} \times 18.02 \, \text{g/mol} = 30.30 \, \text{g} \).

The theoretical yield of water is 30.30 g.

6. **Calculate the percent yield:**
\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 \]
\[ \text{Percent yield} = \left( \frac{18.2 \, \text{g}}{30.30 \, \text{g}} \right) \times 100 \]
\[ \text{Percent yield} = \left( 0.6007 \right) \times 100 \approx 60.1\% \]

Therefore, the percent yield of water is 60.1%, rounded to three significant figures.