40 ml of a hydrocarbon undergoes combustion in 260 ml oxygen and gives 160 mlof `CO_(2)`. If all volumes are measured under similar conditions of temperature and pressue, the formula of the hydrocarbon is

Options

A. `C_(3)H_(8)`
B. `C_(4)H_(8)`
C. `C_(6)H_(14)`
D. `C_(4)H_(10)`

Answer

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

### Step 1: Understand the combustion reaction
The general combustion reaction of a hydrocarbon (CxHy) can be represented as:
\[ C_xH_y + \frac{x + \frac{y}{4}}{1} O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O \]

### Step 2: Write down the volumes given
- Volume of hydrocarbon (CxHy) = 40 ml
- Volume of oxygen (O2) = 260 ml
- Volume of carbon dioxide (CO2) produced = 160 ml

### Step 3: Relate the volumes to moles
Since all volumes are measured under the same conditions of temperature and pressure, we can use the volumes directly to find the number of moles. Thus:
- The volume of CO2 produced (160 ml) corresponds to the number of moles of carbon (x) in the hydrocarbon.
- From the combustion reaction, we know that for every 1 mole of hydrocarbon, x moles of CO2 are produced.

### Step 4: Calculate the value of x
From the volume of CO2 produced:
\[ x = \frac{\text{Volume of CO2}}{\text{Volume of hydrocarbon}} = \frac{160 \text{ ml}}{40 \text{ ml}} = 4 \]
Thus, \( x = 4 \).

### Step 5: Use the volume of oxygen to find y
From the combustion reaction, we can relate the volume of oxygen consumed to the hydrocarbon:
\[ \text{Volume of O2} = 40 \left(4 + \frac{y}{4}\right) \]
Substituting the known volume of O2:
\[ 260 = 40 \left(4 + \frac{y}{4}\right) \]
Simplifying this:
\[ 260 = 160 + 10y \]
\[ 10y = 260 - 160 \]
\[ 10y = 100 \]
\[ y = 10 \]

### Step 6: Write the formula of the hydrocarbon
Now we have \( x = 4 \) and \( y = 10 \). Therefore, the formula of the hydrocarbon is:
\[ C_xH_y = C_4H_{10} \]

### Conclusion
The formula of the hydrocarbon is \( C_4H_{10} \).

---