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 determine the formula of the hydrocarbon from the given data, we can follow these steps:

### Step 1: Write the general combustion reaction for the hydrocarbon
The general formula for the combustion of a hydrocarbon (CxHy) is:
\[ \text{C}_x\text{H}_y + \frac{x + \frac{y}{4}}{O_2} \rightarrow x \text{CO}_2 + \frac{y}{2} \text{H}_2\text{O} \]

### Step 2: Identify the volumes of reactants and products
From the problem:
- Volume of hydrocarbon (C_xH_y) = 10 mL
- Volume of O2 required = 55 mL
- Volume of CO2 produced = 40 mL

### Step 3: Relate the volume of CO2 to the number of moles of carbon (C)
Since 10 mL of hydrocarbon produces 40 mL of CO2, we can relate this to the number of moles of carbon:
- From the reaction, 1 mole of hydrocarbon produces x moles of CO2.
- Therefore, \( 10 \text{ mL of hydrocarbon} \) produces \( 40 \text{ mL of CO2} \), which means:
\[ 10x = 40 \]
\[ x = 4 \]

### Step 4: Relate the volume of O2 to the number of moles of hydrogen (H)
Now, we can use the volume of O2 to find y:
- From the balanced equation, the relationship between O2 and the hydrocarbon is:
\[ 10 \left( x + \frac{y}{4} \right) = 55 \]
Substituting \( x = 4 \):
\[ 10 \left( 4 + \frac{y}{4} \right) = 55 \]
\[ 40 + \frac{10y}{4} = 55 \]
\[ \frac{10y}{4} = 55 - 40 \]
\[ \frac{10y}{4} = 15 \]
\[ 10y = 60 \]
\[ y = 6 \]

### Step 5: Write the empirical formula of the hydrocarbon
Now that we have the values of x and y:
- \( x = 4 \)
- \( y = 6 \)

Thus, the formula of the hydrocarbon is:
\[ \text{C}_4\text{H}_6 \]

### Conclusion
The formula of the hydrocarbon is \( \text{C}_4\text{H}_6 \).

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