A `10 gram` sample of natural gas containing `CH_(4)` and `C_(2)H_(4)` was burnt in excess of oxygen to give `29.0 "grams"` of `CO_(2)` and some water. How many games of water are formed :

Options

A. `9.42 g`
B. `18.81 g`
C. `11.42 g`
D. `15.31 g`

Answer

To solve the problem, we need to determine how many grams of water are formed when a 10 gram sample of natural gas (containing methane \(CH_4\) and ethylene \(C_2H_4\)) is burned, producing 29 grams of carbon dioxide (\(CO_2\)).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equations:**
- For methane (\(CH_4\)):
\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]
- For ethylene (\(C_2H_4\)):
\[ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \]

2. **Combine the Equations:**
- When we combine the two reactions, we get:
\[ CH_4 + C_2H_4 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]
- This shows that 3 moles of \(CO_2\) correspond to 4 moles of \(H_2O\).

3. **Calculate the Molar Masses:**
- Molar mass of \(CO_2\) = \(12 + 2 \times 16 = 44 \, g/mol\)
- Molar mass of \(H_2O\) = \(2 + 16 = 18 \, g/mol\)

4. **Determine the Ratio of Water to Carbon Dioxide:**
- From the balanced equation, we see that:
\[ 3 \, \text{moles of } CO_2 \text{ produce } 4 \, \text{moles of } H_2O \]

5. **Set Up the Proportion:**
- If 3 moles of \(CO_2\) correspond to 4 moles of \(H_2O\), we can set up the following proportion based on the mass of \(CO_2\) produced:
\[ \frac{29 \, g \, CO_2}{3 \times 44 \, g} = \frac{x \, g \, H_2O}{4 \times 18 \, g} \]

6. **Calculate the Mass of Water:**
- Rearranging the equation gives:
\[ x = \frac{29 \, g \times 4 \times 18 \, g}{3 \times 44 \, g} \]
- Now, calculate \(x\):
\[ x = \frac{29 \times 4 \times 18}{3 \times 44} \]
\[ x = \frac{2088}{132} \approx 15.82 \, g \]

### Final Answer:
The mass of water formed is approximately **15.82 grams**.