The combustion of octane can be represented by the equation $$2C_8H_{18}(g) + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g) \quad \Delta H = -10 \ 108 \ kJ \ mol^{-1}$$ The energy produced, in kJ, by the complete oxidation of 45 kg of octane is A - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

First, we need to find the molar mass of octane (C₈H₁₈). The molar mass is:

$ext{Molar mass of C}_8 ext{H}_{18} = (8 \times 12.01) + (18 \times 1.008) = 114.22 \ g \ mol^{-1}$

Next, convert 45 kg to grams:

$45 \ kg = 45000 \ g$

Now, use the formula for moles:

$\text{Moles of octane} = \frac{\text{mass}}{\text{molar mass}} = \frac{45000 \ g}{114.22 \ g \ mol^{-1}} \approx 393.61 \ mol$