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 Answer units

## Step1: Write the balanced chemical equation<br />### The balanced chemical equation for the combustion of methane is:<br />\[ CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O \]<br />## Step2: Calculate moles of methane burned<br />### Molar mass of $CH_{4}$ is $12.01 + 4 \times 1.01 = 16.05$ g/mol. Moles of $CH_{4}$ burned:<br />\[ \text{Moles of } CH_{4} = \frac{1.77 \text{ g}}{16.05 \text{ g/mol}} = 0.1103 \text{ mol} \]<br />## Step3: Determine theoretical yield of water<br />### According to the balanced equation, 1 mole of $CH_{4}$ produces 2 moles of $H_{2}O$. Therefore, moles of $H_{2}O$ produced:<br />\[ \text{Moles of } H_{2}O = 2 \times 0.1103 \text{ mol} = 0.2206 \text{ mol} \]<br />### Molar mass of $H_{2}O$ is $2 \times 1.01 + 16.00 = 18.02$ g/mol. Theoretical yield of $H_{2}O$:<br />\[ \text{Theoretical yield of } H_{2}O = 0.2206 \text{ mol} \times 18.02 \text{ g/mol} = 3.974 \text{ g} \]<br />## Step4: Calculate percent yield<br />### Percent yield is given by:<br />\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 \]<br />### Substituting the values:<br />\[ \text{Percent yield} = \left( \frac{1.79 \text{ g}}{3.974 \text{ g}} \right) \times 100 = 45.04\% \]