A Question 1 (1 point) Retake question Methane burns according to the equation CH_(4)+O_(2)... CO_(2)+H_(2)O You burn 1.77 grams of methane (CH_(4)) in excess oxygen and collect 1.79 grams of water. What is the percent yield for this reaction to the nearest 0.01 ? Your Answer: square square

## Step1: Write the Balanced Chemical Equation<br />### The balanced chemical equation for the combustion of methane is $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$. This shows that one mole of methane produces two moles of water.<br />## Step2: Calculate Moles of Methane Burned<br />### The molar mass of methane ($CH_4$) is $12.01 + 4(1.01) = 16.05 \, \text{g/mol}$. Therefore, the moles of methane burned are $\frac{1.77 \, \text{g}}{16.05 \, \text{g/mol}} = 0.1103 \, \text{mol}$.<br />## Step3: Calculate Theoretical Yield of Water<br />### According to the balanced equation, 1 mole of $CH_4$ produces 2 moles of $H_2O$. Thus, $0.1103 \, \text{mol}$ of $CH_4$ will produce $0.2206 \, \text{mol}$ of $H_2O$. The molar mass of water is $18.02 \, \text{g/mol}$, so the theoretical yield is $0.2206 \, \text{mol} \times 18.02 \, \text{g/mol} = 3.975 \, \text{g}$.<br />## Step4: Calculate Percent Yield<br />### The percent yield is calculated using the formula $\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$. Substituting the values gives $\frac{1.79 \, \text{g}}{3.975 \, \text{g}} \times 100\% = 45.04\%$.