40 mL. of a hydrocarbon undergoes combustion in 260 ml of oxygen and gives 160 ml of carbondioxide. If all gases are measured under similar condtions of temperature and pressure, the formula of hydrocarbon is

Options

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

Explanation

To determine the formula of the hydrocarbon that underwent combustion, we can follow these steps:

### Step 1: Write the general combustion reaction
The general combustion reaction for a hydrocarbon (C_xH_y) can be represented as:
\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Relate volumes of gases to moles
Under the same conditions of temperature and pressure, the volume of gas is directly proportional to the number of moles. Therefore, we can relate the volumes of the reactants and products.

### Step 3: Set up the equations based on the given data
From the problem, we know:
- Volume of hydrocarbon (C_xH_y) = 40 mL
- Volume of oxygen (O_2) = 260 mL
- Volume of carbon dioxide (CO_2) = 160 mL

### Step 4: Determine the value of x (the number of carbon atoms)
From the combustion reaction, we know that:
- 1 mole of hydrocarbon produces x moles of CO_2.
Thus, for 40 mL of hydrocarbon, it produces:
\[ 40 \, \text{mL} \times x = 160 \, \text{mL} \]
Solving for x:
\[ x = \frac{160 \, \text{mL}}{40 \, \text{mL}} = 4 \]

### Step 5: Determine the value of y (the number of hydrogen atoms)
Next, we can set up the equation for the oxygen consumed. The combustion of a hydrocarbon can be expressed as:
\[ C_xH_y + \left( \frac{x + \frac{y}{4}}{1} \right) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O \]

For the volume of oxygen consumed:
\[ 40 \, \text{mL} \left( \frac{4 + \frac{y}{4}}{1} \right) = 260 \, \text{mL} \]

Substituting x = 4 into the equation:
\[ 40 \left( 4 + \frac{y}{4} \right) = 260 \]
\[ 160 + 10y = 260 \]
\[ 10y = 260 - 160 \]
\[ 10y = 100 \]
\[ y = 10 \]

### Step 6: Write the formula of the hydrocarbon
Now that we have the values of x and y:
- x = 4 (number of carbon atoms)
- y = 10 (number of hydrogen atoms)

The formula of the hydrocarbon is:
\[ C_4H_{10} \]

### Conclusion
The hydrocarbon that underwent combustion is butane, with the formula \( C_4H_{10} \).

---