Question

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

CH4 + 2 O2  CO2 + 2 H2O

Answer 1

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

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

From the equation, we can see that 1 mole of CH₄ produces 2 moles of H₂O.

### Step 1: Calculate moles of water produced
First, we need to calculate the number of moles in 15.0 grams of water. The molar mass of water (H₂O) is approximately:

\[
\text{H: 1.01 g/mol} \times 2 + \text{O: 16.00 g/mol} = 2.02 + 16.00 = 18.02 \text{ g/mol}
\]

Now we can calculate the moles of water:

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

### Step 2: Calculate moles of methane needed
According to the balanced equation, 2 moles of H₂O are produced from 1 mole of CH₄. Thus, we can find the moles of CH₄ required:

\[
\text{Moles of CH}_4 = \frac{0.832 \text{ moles of } H_2O}{2} \approx 0.416 \text{ moles of CH}_4
\]

### Step 3: Calculate mass of methane needed
Next, we need to calculate the mass of CH₄. The molar mass of CH₄ is:

\[
\text{C: 12.01 g/mol} + \text{H: 1.01 g/mol} \times 4 = 12.01 + 4.04 = 16.05 \text{ g/mol}
\]

Now calculate the mass of CH₄ needed:

\[
\text{Mass of CH}_4 = \text{moles} \times \text{molar mass} = 0.416 \text{ moles} \times 16.05 \text{ g/mol} \approx 6.6888 \text{ g}
\]

### Conclusion
Therefore, approximately **6.69 grams** of methane (CH₄) are required to produce 15.0 grams of water.