Questions: Gaseous methane (CH₄) will react with gaseous oxygen (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O). Suppose 3.84 g of methane is mixed with 10. g of oxygen. Calculate the maximum mass of methane that could be used by this chemical reaction. Report your answer in g.

Solution Steps

The balanced chemical equation for the combustion of methane is:

\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]

First, calculate the moles of methane (\(\text{CH}_4\)) and oxygen (\(\text{O}_2\)).

• Molar mass of \(\text{CH}_4\) is \(12.01 + 4 \times 1.008 = 16.042 \, \text{g/mol}\).

• Moles of \(\text{CH}_4\) = \(\frac{3.84 \, \text{g}}{16.042 \, \text{g/mol}} = 0.2393 \, \text{mol}\).

• Molar mass of \(\text{O}_2\) is \(2 \times 16.00 = 32.00 \, \text{g/mol}\).

• Moles of \(\text{O}_2\) = \(\frac{10.0 \, \text{g}}{32.00 \, \text{g/mol}} = 0.3125 \, \text{mol}\).

According to the balanced equation, 1 mole of \(\text{CH}_4\) reacts with 2 moles of \(\text{O}_2\). Therefore, the moles of \(\text{O}_2\) required for 0.2393 moles of \(\text{CH}_4\) is:

\[ 0.2393 \, \text{mol} \times 2 = 0.4786 \, \text{mol} \]

Since we only have 0.3125 moles of \(\text{O}_2\), \(\text{O}_2\) is the limiting reactant.

Since \(\text{O}_2\) is the limiting reactant, we calculate the amount of \(\text{CH}_4\) that can react with 0.3125 moles of \(\text{O}_2\):

\[ \text{Moles of } \text{CH}_4 = \frac{0.3125 \, \text{mol}}{2} = 0.15625 \, \text{mol} \]

Convert moles of \(\text{CH}_4\) to grams:

\[ \text{Mass of } \text{CH}_4 = 0.15625 \, \text{mol} \times 16.042 \, \text{g/mol} = 2.5066 \, \text{g} \]

Final Answer