At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of `O_(2)` for complete combustion, and 40 mL of `CO_(2)` is formed. The formula of the hydrocarbon is:

Options

A. `C_(4)H_(6)`
B. `C_(4)H_(7)Cl`
C. `C_(4)H_(10)`
D. `C_(4)H_(8)`

Explanation

To find the formula of the hydrocarbon, we will follow these steps:

### Step 1: Write the general formula for the hydrocarbon
Assume the hydrocarbon has the formula \( C_xH_y \).

### Step 2: Write the balanced combustion reaction
The balanced equation for the combustion of the hydrocarbon can be written as:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 3: Determine the products formed
From the problem, we know that:
- 10 mL of hydrocarbon requires 55 mL of \( O_2 \) for complete combustion.
- 40 mL of \( CO_2 \) is formed.

### Step 4: Relate the volume of \( CO_2 \) to the number of carbon atoms
Since each mole of carbon in the hydrocarbon produces one mole of \( CO_2 \), the volume of \( CO_2 \) produced is directly proportional to the number of carbon atoms \( x \):
\[ 10x = 40 \]
From this, we can solve for \( x \):
\[ x = \frac{40}{10} = 4 \]

### Step 5: Relate the volume of \( O_2 \) to the number of hydrogen atoms
For the complete combustion of the hydrocarbon, the oxygen required can be expressed in terms of \( x \) and \( y \):
The balanced equation gives us:
\[ O_2 \text{ required} = \frac{x + \frac{y}{4}}{10} \]
Setting this equal to the volume of \( O_2 \) used:
\[ \frac{x + \frac{y}{4}}{10} = 55 \]
Substituting \( x = 4 \):
\[ \frac{4 + \frac{y}{4}}{10} = 55 \]

### Step 6: Solve for \( y \)
Multiply both sides by 10:
\[ 4 + \frac{y}{4} = 550 \]
Subtract 4 from both sides:
\[ \frac{y}{4} = 546 \]
Multiply by 4:
\[ y = 546 \times 4 = 6 \]

### Step 7: Write the final formula of the hydrocarbon
Now we have \( x = 4 \) and \( y = 6 \), so the formula of the hydrocarbon is:
\[ C_4H_6 \]

### Conclusion
The formula of the hydrocarbon is \( C_4H_6 \).

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