To solve the problem of calculating the volume of \( O_2 \) required for the complete combustion of 2.64 L of acetylene \( (C_2H_2) \) at 1 atm and 273 K, we will follow these steps:
### Step 1: Write the Balanced Chemical Equation
The balanced equation for the combustion of acetylene is:
\[
2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(l)
\]
### Step 2: Calculate the Moles of Acetylene
Using the ideal gas law, we can calculate the number of moles of acetylene \( (C_2H_2) \):
\[
n = \frac{PV}{RT}
\]
Where:
- \( P = 1 \, \text{atm} \)
- \( V = 2.64 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 273 \, \text{K} \)
Substituting the values:
\[
n_{C_2H_2} = \frac{(1 \, \text{atm})(2.64 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(273 \, \text{K})}
\]
Calculating this gives:
\[
n_{C_2H_2} \approx 0.118 \, \text{moles}
\]
### Step 3: Determine the Moles of Oxygen Required
From the balanced equation, we see that 2 moles of \( C_2H_2 \) react with 5 moles of \( O_2 \). Therefore, the ratio of \( O_2 \) to \( C_2H_2 \) is:
\[
\frac{5 \, \text{moles } O_2}{2 \, \text{moles } C_2H_2}
\]
To find the moles of \( O_2 \) needed:
\[
n_{O_2} = n_{C_2H_2} \times \frac{5}{2}
\]
Substituting the value of \( n_{C_2H_2} \):
\[
n_{O_2} = 0.118 \times \frac{5}{2} = 0.294 \, \text{moles}
\]
### Step 4: Calculate the Volume of Oxygen Required
Now we calculate the volume of \( O_2 \) using the ideal gas law again:
\[
V = \frac{nRT}{P}
\]
Substituting the values:
\[
V_{O_2} = \frac{(0.294 \, \text{moles})(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(273 \, \text{K})}{1 \, \text{atm}}
\]
Calculating this gives:
\[
V_{O_2} \approx 6.589 \, \text{L}
\]
Rounding this to two decimal places, we get:
\[
V_{O_2} \approx 6.6 \, \text{L}
\]
### Final Answer
The volume of \( O_2 \) required for the complete combustion of 2.64 L of acetylene at 1 atm and 273 K is approximately **6.6 liters**.
---
