Certainly! Let's go through a detailed, step-by-step solution to determine the heat of reaction when 3.00 grams of octane ([tex]$C_8H_{18}$[/tex]) burns in a bomb calorimeter, given that the enthalpy of combustion ([tex]$\Delta H_{\text{comb}}$[/tex]) is [tex]\(-5461 \text{kJ/mol}\)[/tex].
1. Determine the Molar Mass of Octane:
The molecular formula for octane is [tex]$C_8H_{18}$[/tex]. To find the molar mass, we calculate the total mass of all atoms in the molecule:
[tex]\[
\text{Molar mass of } C_8H_{18} = 8 \text{carbon atoms} \times 12.01 \text{g/mol} + 18 \text{hydrogen atoms} \times 1.01 \text{g/mol}
\][/tex]
[tex]\[
= (8 \times 12.01) + (18 \times 1.01)
\][/tex]
[tex]\[
= 96.08 + 18.18
\][/tex]
[tex]\[
= 114.26 \text{g/mol}
\][/tex]
2. Calculate the Moles of Octane Burned:
Using the molar mass, we can determine how many moles of octane are in 3.00 grams:
[tex]\[
\text{Moles of } C_8H_{18} = \frac{\text{mass of } C_8H_{18}}{\text{molar mass of } C_8H_{18}}
\][/tex]
[tex]\[
= \frac{3.00 \text{g}}{114.26 \text{g/mol}}
\][/tex]
[tex]\[
= 0.026256 \text{ moles} \quad \text{(keeping significant figures)}
\][/tex]
3. Calculate the Heat of Reaction ([tex]\(q_{\text{comb}}\)[/tex]):
Given the enthalpy of combustion [tex]\(\Delta H_{\text{comb}} = -5461 \text{kJ/mol}\)[/tex], we find the heat released or absorbed during the combustion of the calculated moles of octane.
[tex]\[
q_{\text{comb}} = \text{moles of } C_8H_{18} \times \Delta H_{\text{comb}}
\][/tex]
[tex]\[
= 0.026256 \text{ moles} \times (-5461 \text{kJ/mol})
\][/tex]
[tex]\[
= -143.384 \text{kJ}
\][/tex]
Since we need to report the answer with the correct significant figures, we reflect the significant figures from the input data (3.00 g). Thus, the final answer, emphasizing both the sign and magnitude:
The heat of reaction when 3.00 grams of [tex]$C_8H_{18}$[/tex] burns is:
[tex]\[
-143.38 \text{kJ}
\][/tex]
This solution takes into account the significant figures and the correct methodology for calculating the heat of reaction.
