First, we need to find the molar mass of methane (CH\(_4\)):
\[
\text{Molar mass of CH}_4 = 12.01 \, (\text{C}) + 4 \times 1.008 \, (\text{H}) = 16.042 \, \text{g/mol}
\]
Next, we calculate the moles of methane:
\[
\text{Moles of CH}_4 = \frac{8.32 \, \text{g}}{16.042 \, \text{g/mol}} = 0.5185 \, \text{mol}
\]
From the balanced chemical equation, we see that 1 mole of CH\(_4\) reacts with 2 moles of O\(_2\):
\[
\text{Moles of O}_2 = 0.5185 \, \text{mol CH}_4 \times 2 \, \frac{\text{mol O}_2}{\text{mol CH}_4} = 1.037 \, \text{mol O}_2
\]
Now, we find the molar mass of O\(_2\):
\[
\text{Molar mass of O}_2 = 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol}
\]
Then, we calculate the mass of O\(_2\) used:
\[
\text{Mass of O}_2 = 1.037 \, \text{mol} \times 32.00 \, \text{g/mol} = 33.18 \, \text{g}
\]
From the balanced chemical equation, 1 mole of CH\(_4\) produces 1 mole of CO\(_2\) and 2 moles of H\(_2\)O:
\[
\text{Moles of CO}_2 = 0.5185 \, \text{mol CH}_4 \times 1 \, \frac{\text{mol CO}_2}{\text{mol CH}_4} = 0.5185 \, \text{mol CO}_2
\]
\[
\text{Moles of H}_2\text{O} = 0.5185 \, \text{mol CH}_4 \times 2 \, \frac{\text{mol H}_2\text{O}}{\text{mol CH}_4} = 1.037 \, \text{mol H}_2\text{O}
\]
First, we find the molar masses of CO\(_2\) and H\(_2\)O:
\[
\text{Molar mass of CO}_2 = 12.01 \, (\text{C}) + 2 \times 16.00 \, (\text{O}) = 44.01 \, \text{g/mol}
\]
\[
\text{Molar mass of H}_2\text{O} = 2 \times 1.008 \, (\text{H}) + 16.00 \, (\text{O}) = 18.016 \, \text{g/mol}
\]
Then, we calculate the masses:
\[
\text{Mass of CO}_2 = 0.5185 \, \text{mol} \times 44.01 \, \text{g/mol} = 22.81 \, \text{g}
\]
\[
\text{Mass of H}_2\text{O} = 1.037 \, \text{mol} \times 18.016 \, \text{g/mol} = 18.68 \, \text{g}
\]
\[
\boxed{\text{Mass of O}_2 = 33.18 \, \text{g}}
\]
\[
\boxed{\text{Mass of CO}_2 = 22.81 \, \text{g}}
\]
\[
\boxed{\text{Mass of H}_2\text{O} = 18.68 \, \text{g}}
\]
Answered
