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Using the following equation for the combustion of octane, calculate the heat associated with the combustion of 150.0 g of octane, assuming complete combustion. The molar mass of octane is [tex]114.33 \, \text{g/mole}[/tex]. The molar mass of oxygen is [tex]31.9988 \, \text{g/mole}[/tex].

[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]

A. [tex]-803.1 \, \text{kJ}[/tex]
B. [tex]-7228 \, \text{kJ}[/tex]
C. [tex]-903.4 \, \text{kJ}[/tex]
D. [tex]-578.2 \, \text{kJ}[/tex]
E. [tex]-11018 \, \text{kJ}[/tex]



Answer :

GinnyAnswer
To solve this problem, let's break it down step by step:

### 1. Identify the Given Data:
- Mass of octane ([tex]\(C_8H_{18}\)[/tex]): 150.0 g
- Molar mass of octane: 114.33 g/mole
- Enthalpy change ([tex]\(\Delta H_{\text{rxn}}\)[/tex]): -11818 kJ for the reaction involving 2 moles of octane.

### 2. Determine the Number of Moles of Octane:
To find the number of moles of octane, we use the formula:
[tex]\[ \text{moles of octane} = \frac{\text{mass of octane}}{\text{molar mass of octane}} \][/tex]

Substituting the given values:
[tex]\[ \text{moles of octane} = \frac{150.0 \text{ g}}{114.33 \text{ g/mole}} \approx 1.311991603253739 \text{ moles} \][/tex]

### 3. Calculate the Heat Associated with Combustion:
From the balanced chemical equation, we know that the enthalpy change [tex]\(\Delta H_{\text{rxn}}\)[/tex] of -11818 kJ is for 2 moles of octane. To find the heat associated with the combustion of the calculated moles of octane, we use the proportional relationship.

First, we find the enthalpy change per mole of octane:
[tex]\[ \Delta H_{\text{rxn}} \text{ per mole of octane} = \frac{\Delta H_{\text{rxn}}}{2} = \frac{-11818 \text{ kJ}}{2} = -5909 \text{ kJ/mole} \][/tex]

Now, we calculate the total heat associated with the combustion of the calculated moles of octane:
[tex]\[ \text{Heat combustion} = (\text{moles of octane}) \times (\Delta H_{\text{rxn}} \text{ per mole of octane}) \][/tex]

Substituting the values:
[tex]\[ \text{Heat combustion} = 1.311991603253739 \text{ moles} \times (-5909 \text{ kJ/mole}) \approx -7752.558383626344 \text{ kJ} \][/tex]

### Final Answer:
The heat associated with the combustion of 150.0 g of octane is approximately [tex]\(-7752.56 \text{ kJ}\)[/tex].

Among the given options:
-803.1 kJ
-7228 kJ
-903.4 kJ
-578.2 kJ
-11018 kJ
None of these options match the calculated result directly. However, the detailed calculation shows that the combustion of 150.0 g of octane releases approximately [tex]\(-7752.56 \text{ kJ}\)[/tex].