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. 4 L
D. 84 L

Explanation

To solve the problem of how many liters of \( O_2 \) at STP are needed to complete the combustion of 39 g of liquid benzene \( C_6H_6 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the combustion of benzene is:
\[ 2C_6H_6 (l) + 15O_2 (g) \rightarrow 12CO_2 (g) + 6H_2O (g) \]

### Step 2: Calculate the number of moles of benzene
To find the number of moles of benzene in 39 g, we use the molar mass of benzene, which is 78 g/mol.

\[ \text{Moles of } C_6H_6 = \frac{\text{mass}}{\text{molar mass}} = \frac{39 \, \text{g}}{78 \, \text{g/mol}} = 0.5 \, \text{moles} \]

### Step 3: Use the stoichiometry of the reaction
From the balanced equation, we see that 2 moles of benzene react with 15 moles of \( O_2 \). Therefore, we can set up a ratio to find the moles of \( O_2 \) needed for 0.5 moles of benzene.

\[ \text{Moles of } O_2 = 0.5 \, \text{moles } C_6H_6 \times \frac{15 \, \text{moles } O_2}{2 \, \text{moles } C_6H_6} = 3.75 \, \text{moles } O_2 \]

### Step 4: Convert moles of \( O_2 \) to grams
Now, we can convert the moles of \( O_2 \) to grams using the molar mass of \( O_2 \), which is 32 g/mol.

\[ \text{Mass of } O_2 = 3.75 \, \text{moles} \times 32 \, \text{g/mol} = 120 \, \text{g} \]

### Step 5: Convert grams of \( O_2 \) to liters at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, we can convert the moles of \( O_2 \) to liters.

\[ \text{Liters of } O_2 = 3.75 \, \text{moles} \times 22.4 \, \text{L/mol} = 84 \, \text{L} \]

### Final Answer
Thus, the volume of \( O_2 \) needed to complete the combustion of 39 g of liquid benzene is **84 liters**.

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