To calculate the heat associated with the combustion of 150.8 g of octane given the reaction:
[tex]\[ 2 \, \text{C}_8\text{H}_{18} + 25 \, \text{O}_2 \rightarrow 16 \, \text{CO}_2 + 18 \, \text{H}_2\text{O} \quad \Delta H ^{\circ}_{\text{rxn}} = -11818 \, \text{kJ} \][/tex]
we need to follow these steps:
1. Determine the moles of octane (C[tex]\(_8\)[/tex]H[tex]\(_{18}\)[/tex]):
The molar mass of octane ([tex]\(\text{C}_8\text{H}_{18}\)[/tex]) is [tex]\(114.33 \, \text{g/mol}\)[/tex]. To find the number of moles, use the formula:
[tex]\[
\text{moles of octane} = \frac{\text{mass of octane}}{\text{molar mass of octane}}
\][/tex]
Given the mass of octane is [tex]\(150.8 \, \text{g}\)[/tex],
[tex]\[
\text{moles of octane} = \frac{150.8 \, \text{g}}{114.33 \, \text{g/mol}} = 1.318988891804426 \, \text{moles}
\][/tex]
2. Relate the moles of octane to the heat of reaction:
The given reaction shows that [tex]\(\Delta H ^{\circ}_{\text{rxn}} = -11818 \, \text{kJ}\)[/tex] for 2 moles of octane combusted.
Thus, for 1 mole of octane, the heat released would be:
[tex]\[
\text{Heat for 1 mole of octane} = \frac{-11818 \, \text{kJ}}{2 \, \text{moles}} = -5909 \, \text{kJ/mole}
\][/tex]
3. Calculate the heat associated with the combustion of the given moles of octane:
Multiply the moles of octane by the heat released per mole.
[tex]\[
\text{Heat combustion} = 1.318988891804426 \, \text{moles} \times -5909 \, \text{kJ/mole}
\][/tex]
[tex]\[
\text{Heat combustion} = -7793.905361672353 \, \text{kJ}
\][/tex]
Therefore, the heat associated with the complete combustion of 150.8 g of octane is [tex]\(-7793.905361672353 \, \text{kJ}\)[/tex].
