Liquid benzene `C_(6)H_(6))` burns in oxygen according to the equation, `2C_(6)H_(6)(l)+15O_(2)(g)to12CO_(2)(g)+6H_(2)O(g)`
How many litres of `O_(2)` at STP are needed to complete the combustion of 39 g of liquid benzene ? (Mol . Weight if `O_(2)=32,C_(6)H_(6)=78)`

Options

A. 74 L
B. 11.2 L
C. 22.4L
D. 84 L

Answer

To solve the problem, we need to determine how many liters of oxygen (O₂) at standard temperature and pressure (STP) are required to completely combust 39 grams of liquid benzene (C₆H₆).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The combustion of benzene is represented by the following balanced equation:
\[ 2C_6H_6(l) + 15O_2(g) \rightarrow 12CO_2(g) + 6H_2O(g) \]

2. **Calculate the Molar Mass of Benzene (C₆H₆)**:
The molar mass of benzene (C₆H₆) is given as 78 g/mol.

3. **Determine Moles of Benzene**:
To find the number of moles of benzene in 39 grams, use the formula:
\[ \text{Moles of } C_6H_6 = \frac{\text{mass}}{\text{molar mass}} = \frac{39 \text{ g}}{78 \text{ g/mol}} = 0.5 \text{ moles} \]

4. **Use Stoichiometry to Find Moles of O₂ Required**:
From the balanced equation, 2 moles of C₆H₆ react with 15 moles of O₂. Therefore, the moles of O₂ required for 0.5 moles of C₆H₆ can be calculated as follows:
\[ \text{Moles of } O_2 = 0.5 \text{ moles } C_6H_6 \times \frac{15 \text{ moles } O_2}{2 \text{ moles } C_6H_6} = 3.75 \text{ moles } O_2 \]

5. **Calculate the Mass of O₂ Required**:
The molar mass of O₂ is given as 32 g/mol. Therefore, the mass of O₂ required is:
\[ \text{Mass of } O_2 = \text{moles} \times \text{molar mass} = 3.75 \text{ moles} \times 32 \text{ g/mol} = 120 \text{ g} \]

6. **Convert Mass of O₂ to Volume at STP**:
At STP, 1 mole of any gas occupies 22.4 liters. Therefore, the volume of O₂ required can be calculated as:
\[ \text{Volume of } O_2 = \text{moles} \times 22.4 \text{ L/mol} = 3.75 \text{ moles} \times 22.4 \text{ L/mol} = 84 \text{ L} \]

### Final Answer:
The volume of O₂ required at STP to combust 39 grams of liquid benzene is **84 liters**.