Question

Use the chemical equation to answer the question. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) The molar mass of carbon (C) is 12 g/mol, the molar mass of hydrogen (H) is 1 g/mol, and the molar mass of oxygen (O) is 16 g/mol. When 48 grams of methane (CH4) react completely with oxygen gas (O2), how many grams of water are produced in the reaction?

Answer 1

To solve this problem, we first need to determine the number of moles of methane (CH₄) in the given mass (48 grams), and then use the balanced chemical equation to find the amount of water (H₂O) produced.

1. **Calculate the molar mass of methane (CH₄)**:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol × 4 = 4 g/mol
- Molar mass of CH₄ = 12 g/mol + 4 g/mol = 16 g/mol

2. **Calculate the number of moles of methane (CH₄)**:
\[
\text{Moles of CH₄} = \frac{\text{mass}}{\text{molar mass}} = \frac{48 \text{ g}}{16 \text{ g/mol}} = 3 \text{ moles of CH₄}
\]

3. **Use the balanced equation to find the relationship between CH₄ and H₂O**:
The balanced reaction is:
\[
\text{CH₄(g)} + 2\text{O₂(g)} \rightarrow \text{CO₂(g)} + 2\text{H₂O(g)}
\]
From the equation, 1 mole of CH₄ produces 2 moles of H₂O.

4. **Calculate the moles of water (H₂O) produced**:
\[
\text{Moles of H₂O} = 3 \text{ moles of CH₄} \times 2 \text{ moles of H₂O/mole of CH₄} = 6 \text{ moles of H₂O}
\]

5. **Calculate the molar mass of water (H₂O)**:
- Hydrogen (H) = 1 g/mol × 2 = 2 g/mol
- Oxygen (O) = 16 g/mol
- Molar mass of H₂O = 2 g/mol + 16 g/mol = 18 g/mol

6. **Calculate the grams of water produced**:
\[
\text{Mass of H₂O} = \text{moles} \times \text{molar mass} = 6 \text{ moles} \times 18 \text{ g/mol} = 108 \text{ g}
\]

Therefore, when 48 grams of methane react completely with oxygen, \(108\) grams of water are produced.