How much energy is released when 6 mole of octane is burnt in air ? Given `DeltaH_(f)^(@)` for `CO_(2)(g),H_(2)O(g)` and `C_(8)H_(18)(l)` respectively are `-490,-240` and `+160J//mol`

Options

A. `-6.2 KJ`
B. `-37.4 KJ`
C. `-35.5 KJ`
D. `-20.0 KJ.`

Explanation

To determine the energy released when 6 moles of octane (C₈H₁₈) are burned in air, we can use the standard enthalpy of formation (ΔH_f) values provided for the products (CO₂ and H₂O) and the reactant (C₈H₁₈).

### Step-by-Step Solution:

1. **Write the Balanced Combustion Reaction for Octane:**
The balanced equation for the combustion of octane is:
\[ 2 \text{C}_8\text{H}_{18}(l) + 25 \text{O}_2(g) \rightarrow 16 \text{CO}_2(g) + 18 \text{H}_2\text{O}(g) \]

2. **Identify the ΔH_f Values:**
- ΔH_f for CO₂(g) = -490 kJ/mol
- ΔH_f for H₂O(g) = -240 kJ/mol
- ΔH_f for C₈H₁₈(l) = +160 kJ/mol

3. **Calculate the Total Enthalpy Change (ΔH) for the Reaction:**
The enthalpy change for the reaction can be calculated using the formula:
\[ \Delta H = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)} \]

For the products:
- For 16 moles of CO₂: \(16 \times (-490) = -7840 \text{ kJ}\)
- For 18 moles of H₂O: \(18 \times (-240) = -4320 \text{ kJ}\)

Total for products:
\[ -7840 + (-4320) = -12160 \text{ kJ} \]

For the reactants:
- For 2 moles of C₈H₁₈: \(2 \times 160 = +320 \text{ kJ}\)

Total for reactants:
\[ +320 \text{ kJ} \]

Now, substituting into the ΔH formula:
\[ \Delta H = (-12160) - (+320) = -12480 \text{ kJ} \]

4. **Calculate the Energy Released for 6 Moles of Octane:**
The ΔH calculated is for 2 moles of octane. To find the energy released for 6 moles:
\[ \Delta H \text{ for 6 moles} = \left(\frac{-12480 \text{ kJ}}{2}\right) \times 6 = -37440 \text{ kJ} \]

5. **Convert to Kilojoules:**
Since the energy released is already in kilojoules, we can express it as:
\[ -37.440 \text{ kJ} \]

### Final Answer:
The energy released when 6 moles of octane are burned in air is approximately **-37.4 kJ**.