Assuming that petrol is octane `(C_(8)H_(18))` and has density `0.8 g//ml, 1.425 litre` of petrol on complete combustion will consume

Options

A. (a)`50` mole of `O_(2)`
B. (b)`125` mole of `O_(2)`
C. (c )`100` mole of `O_(2)`
D. (d)`200` mole of `O_(2)`

Explanation

To solve the problem, we need to determine how much oxygen (O₂) is required for the complete combustion of 1.425 liters of petrol, assuming it is octane (C₈H₁₈) with a density of 0.8 g/ml.

### Step-by-Step Solution:

1. **Calculate the Mass of Octane:**
- Given the density of octane is 0.8 g/ml and the volume is 1.425 liters.
- Convert liters to milliliters:
\[ 1.425 \text{ L} = 1.425 \times 1000 \text{ ml} = 1425 \text{ ml} \]
- Use the formula for mass:
\[ \text{Mass} = \text{Density} \times \text{Volume} = 0.8 \text{ g/ml} \times 1425 \text{ ml} = 1140 \text{ g} \]

2. **Calculate the Number of Moles of Octane:**
- The molecular mass of octane (C₈H₁₈) can be calculated as follows:
- Carbon (C) has an atomic mass of 12 g/mol, and there are 8 carbon atoms:
\[ 8 \times 12 = 96 \text{ g/mol} \]
- Hydrogen (H) has an atomic mass of 1 g/mol, and there are 18 hydrogen atoms:
\[ 18 \times 1 = 18 \text{ g/mol} \]
- Therefore, the molecular mass of octane:
\[ \text{Molecular mass} = 96 + 18 = 114 \text{ g/mol} \]
- Now, calculate the number of moles of octane:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molecular mass}} = \frac{1140 \text{ g}}{114 \text{ g/mol}} = 10 \text{ moles} \]

3. **Determine the Oxygen Requirement for Combustion:**
- The balanced equation for the combustion of octane is:
\[ C_8H_{18} + \frac{25}{2} O_2 \rightarrow 8 CO_2 + 9 H_2O \]
- From the equation, 1 mole of octane requires \(\frac{25}{2}\) moles of O₂.
- For 10 moles of octane:
\[ \text{Oxygen required} = 10 \text{ moles of } C_8H_{18} \times \frac{25}{2} \text{ moles of } O_2 = 125 \text{ moles of } O_2 \]

### Final Answer:
The amount of oxygen required for the complete combustion of 1.425 liters of petrol (octane) is **125 moles of O₂**.

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