What is the volume of `O_(2)` needed to burn 60 L of octane at STP ?
Equaton `2C_(6)H_(8)+25O_(2) to 16CO_(2)+18H_(2)O`
R.A.N.:C=12,H=1,O=16

Explanation

To find the volume of \( O_2 \) needed to burn 60 L of octane (\( C_8H_{18} \)) at STP, we will follow these steps:

### Step 1: Write the balanced chemical equation
The complete balanced reaction for the combustion of octane is:
\[ 2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O \]

### Step 2: Determine the volume of octane in moles
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. However, since octane is a liquid, we will need to convert the volume of octane to moles using its molar mass.

1. **Calculate the molar mass of octane (\( C_8H_{18} \))**:
- Carbon (C) = 12 g/mol, Hydrogen (H) = 1 g/mol
- Molar mass of octane = \( (8 \times 12) + (18 \times 1) = 96 + 18 = 114 \) g/mol

2. **Convert the volume of octane to mass**:
- Given volume of octane = 60 L
- Since the density of octane is approximately 0.703 g/mL, we convert liters to grams:
\[ \text{Mass of octane} = 60 \, \text{L} \times 1000 \, \text{mL/L} \times 0.703 \, \text{g/mL} = 42180 \, \text{g} \]

3. **Calculate the number of moles of octane**:
\[ \text{Number of moles of octane} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{42180 \, \text{g}}{114 \, \text{g/mol}} \approx 370.53 \, \text{moles} \]

### Step 3: Use stoichiometry to find the volume of \( O_2 \) required
From the balanced equation:
- 2 moles of \( C_8H_{18} \) require 25 moles of \( O_2 \).

1. **Calculate the moles of \( O_2 \) needed for the moles of octane**:
\[ \text{Moles of } O_2 = \left( \frac{25 \, \text{moles of } O_2}{2 \, \text{moles of } C_8H_{18}} \right) \times 370.53 \, \text{moles of } C_8H_{18} \approx 4626.63 \, \text{moles of } O_2 \]

2. **Convert moles of \( O_2 \) to volume**:
\[ \text{Volume of } O_2 = \text{Moles} \times 22.4 \, \text{L/mol} = 4626.63 \, \text{moles} \times 22.4 \, \text{L/mol} \approx 103,000 \, \text{L} \]

### Final Answer
The volume of \( O_2 \) needed to burn 60 L of octane at STP is approximately **103,000 L**.

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