The formula of a gaseous hydrocarbon which requires 6 times of its own volume of `O_2` for complete oxidation and produces 4 time its own volume of `CO_2` is `C_(x)H_(y)`. The value of y is _________

Answer

To solve the problem, we need to determine the value of \( y \) in the hydrocarbon formula \( C_xH_y \) based on the information provided about the volumes of oxygen and carbon dioxide produced during combustion.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The complete combustion of a hydrocarbon \( C_xH_y \) can be represented by the following balanced chemical equation:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

2. **Volume Relationships**:
According to the problem:
- The hydrocarbon requires 6 times its own volume of \( O_2 \) for complete oxidation.
- It produces 4 times its own volume of \( CO_2 \).

Let the volume of the hydrocarbon be \( V \). Then:
- Volume of \( O_2 \) required = \( 6V \)
- Volume of \( CO_2 \) produced = \( 4V \)

3. **Using Stoichiometry**:
From the balanced equation, we can derive the stoichiometric relationships:
- For every 1 mole of \( C_xH_y \), it reacts with \( \frac{x+y}{4} \) moles of \( O_2 \) and produces \( x \) moles of \( CO_2 \) and \( \frac{y}{2} \) moles of \( H_2O \).

Therefore, the relationship between the volumes can be expressed as:
\[ \text{Volume of } O_2 = \frac{x+y}{4} \times V \]
\[ \text{Volume of } CO_2 = x \times V \]

4. **Setting Up Equations**:
From the volume relationships:
- For \( O_2 \):
\[ 6V = \frac{x+y}{4} \times V \]
Dividing both sides by \( V \) (assuming \( V \neq 0 \)):
\[ 6 = \frac{x+y}{4} \]
Multiplying through by 4:
\[ x + y = 24 \quad \text{(Equation 1)} \]

- For \( CO_2 \):
\[ 4V = x \times V \]
Dividing both sides by \( V \):
\[ 4 = x \quad \text{(Equation 2)} \]

5. **Substituting to Find \( y \)**:
Now, substitute \( x = 4 \) from Equation 2 into Equation 1:
\[ 4 + y = 24 \]
Solving for \( y \):
\[ y = 24 - 4 = 20 \]

6. **Final Answer**:
The value of \( y \) is \( \boxed{20} \).