Answer :
### Step-by-Step Solution
Given Reaction:
[tex]\[C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l) \quad \Delta H = -2220 \text{ kJ}\][/tex]
Step 1: Determine the Type of Reaction
- The given [tex]\(\Delta H\)[/tex] for the reaction is [tex]\(-2220 \text{ kJ}\)[/tex].
- Since [tex]\(\Delta H\)[/tex] is negative, the reaction releases energy.
- Therefore, the reaction is exothermic.
Step 2: Calculate the Molar Mass of [tex]\(C_3H_8\)[/tex]
- Calculate the molar masses of Carbon (C) and Hydrogen (H):
- Atomic mass of Carbon (C): 12.01 g/mol
- Atomic mass of Hydrogen (H): 1.01 g/mol
- Molar mass of [tex]\(C_3H_8\)[/tex] (Propane):
[tex]\[ \text{Molar mass of } C_3H_8 = 3 \times 12.01 \text{ g/mol} + 8 \times 1.01 \text{ g/mol} = 36.03 \text{ g/mol} + 8.08 \text{ g/mol} = 44.11 \text{ g/mol} \][/tex]
Step 3: Calculate the Number of Moles of [tex]\(C_3H_8\)[/tex]
- Given mass of [tex]\(C_3H_8\)[/tex] is 86.1 g.
- Number of moles of [tex]\(C_3H_8\)[/tex]:
[tex]\[ \text{Moles of } C_3H_8 = \frac{\text{mass of } C_3H_8}{\text{molar mass of } C_3H_8} = \frac{86.1 \text{ g}}{44.11 \text{ g/mol}} = 1.951938335978236 \text{ moles} \][/tex]
Step 4: Calculate the Heat Released
- The heat released ([tex]\(\Delta H\)[/tex]) per mole of [tex]\(C_3H_8\)[/tex] is [tex]\(-2220 \text{ kJ}\)[/tex].
- Thus, for 1.951938335978236 moles of [tex]\(C_3H_8\)[/tex], the heat released is:
[tex]\[ \text{Heat released} = \text{moles of } C_3H_8 \times \Delta H = 1.951938335978236 \text{ moles} \times (-2220 \text{ kJ/mol}) = -4333.303105871684 \text{ kJ} \][/tex]
### Summary
[tex]\[ \begin{array}{|c|c|} \hline \text{This reaction is...} & \text{exothermic} \\ \hline \end{array} \][/tex]
When 86.1 g of [tex]\(C_3H_8\)[/tex] react, the reaction will release heat.
[tex]\[ \begin{array}{|c|c|} \hline \text{Will any heat be released or absorbed?} & \text{Yes, heat is released} \\ \hline \end{array} \][/tex]
### Conclusion
- The reaction is exothermic.
- When 86.1 g of [tex]\(C_3H_8\)[/tex] react, approximately [tex]\(-4333.3 \text{ kJ}\)[/tex] of heat will be released.
Given Reaction:
[tex]\[C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l) \quad \Delta H = -2220 \text{ kJ}\][/tex]
Step 1: Determine the Type of Reaction
- The given [tex]\(\Delta H\)[/tex] for the reaction is [tex]\(-2220 \text{ kJ}\)[/tex].
- Since [tex]\(\Delta H\)[/tex] is negative, the reaction releases energy.
- Therefore, the reaction is exothermic.
Step 2: Calculate the Molar Mass of [tex]\(C_3H_8\)[/tex]
- Calculate the molar masses of Carbon (C) and Hydrogen (H):
- Atomic mass of Carbon (C): 12.01 g/mol
- Atomic mass of Hydrogen (H): 1.01 g/mol
- Molar mass of [tex]\(C_3H_8\)[/tex] (Propane):
[tex]\[ \text{Molar mass of } C_3H_8 = 3 \times 12.01 \text{ g/mol} + 8 \times 1.01 \text{ g/mol} = 36.03 \text{ g/mol} + 8.08 \text{ g/mol} = 44.11 \text{ g/mol} \][/tex]
Step 3: Calculate the Number of Moles of [tex]\(C_3H_8\)[/tex]
- Given mass of [tex]\(C_3H_8\)[/tex] is 86.1 g.
- Number of moles of [tex]\(C_3H_8\)[/tex]:
[tex]\[ \text{Moles of } C_3H_8 = \frac{\text{mass of } C_3H_8}{\text{molar mass of } C_3H_8} = \frac{86.1 \text{ g}}{44.11 \text{ g/mol}} = 1.951938335978236 \text{ moles} \][/tex]
Step 4: Calculate the Heat Released
- The heat released ([tex]\(\Delta H\)[/tex]) per mole of [tex]\(C_3H_8\)[/tex] is [tex]\(-2220 \text{ kJ}\)[/tex].
- Thus, for 1.951938335978236 moles of [tex]\(C_3H_8\)[/tex], the heat released is:
[tex]\[ \text{Heat released} = \text{moles of } C_3H_8 \times \Delta H = 1.951938335978236 \text{ moles} \times (-2220 \text{ kJ/mol}) = -4333.303105871684 \text{ kJ} \][/tex]
### Summary
[tex]\[ \begin{array}{|c|c|} \hline \text{This reaction is...} & \text{exothermic} \\ \hline \end{array} \][/tex]
When 86.1 g of [tex]\(C_3H_8\)[/tex] react, the reaction will release heat.
[tex]\[ \begin{array}{|c|c|} \hline \text{Will any heat be released or absorbed?} & \text{Yes, heat is released} \\ \hline \end{array} \][/tex]
### Conclusion
- The reaction is exothermic.
- When 86.1 g of [tex]\(C_3H_8\)[/tex] react, approximately [tex]\(-4333.3 \text{ kJ}\)[/tex] of heat will be released.
