Question

How many grams of methane (CH4) are required to make 15.0 grams of water?

CH4 + 2 O2  CO2 + 2 H2O
Responses

88.75 g
88.75 g

6.67 g
6.67 g

0.3 g
0.3 g

322 g
322 g

Answer 1

To determine how many grams of methane (CH₄) are required to produce 15.0 grams of water (H₂O), we need to use stoichiometry based on the balanced chemical equation:

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

From the equation, 1 mole of methane produces 2 moles of water.

**Step 1: Calculate the moles of water produced.**

The molar mass of water (H₂O) is approximately 18.02 g/mol.

\[
\text{Moles of } H_2O = \frac{\text{mass}}{\text{molar mass}} = \frac{15.0 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.832 \, \text{mol}
\]

**Step 2: Find the moles of CH₄ needed.**

From the balanced equation, 1 mole of CH₄ produces 2 moles of H₂O. Thus:

\[
\text{Moles of CH}_4 = \frac{0.832 \, \text{mol H}_2O}{2} = 0.416 \, \text{mol CH}_4
\]

**Step 3: Convert moles of CH₄ to grams.**

The molar mass of methane (CH₄) is approximately 16.04 g/mol.

\[
\text{Mass of CH}_4 = \text{moles} \times \text{molar mass} = 0.416 \, \text{mol} \times 16.04 \, \text{g/mol} \approx 6.67 \, \text{g}
\]

Thus, the amount of methane required to produce 15.0 grams of water is approximately **6.67 grams**.

The correct answer is:

**6.67 g**