Sure, let's go through the solution step-by-step to determine how many grams of rust (Fe₂O₃) form when 230 kJ of energy are released.
### Step 1: Determine the moles of Fe₂O₃ formed per kJ of energy released
We are given the enthalpy change ([tex]\(\Delta H\)[/tex]) for the formation of Fe₂O₃:
[tex]\[ \Delta H = -1.7 \times 10^3 \text{ kJ} \][/tex]
This enthalpy change corresponds to the formation of 2 moles of Fe₂O₃. Thus, the moles of Fe₂O₃ formed per kJ of energy can be calculated as:
[tex]\[ \text{Moles of Fe₂O₃ per kJ} = \frac{2 \text{ moles}}{-1.7 \times 10^3 \text{ kJ}} = -0.001176470588235294 \text{ moles per kJ} \][/tex]
### Step 2: Calculate the moles of Fe₂O₃ formed when 230 kJ of energy are released
Given:
[tex]\[ q_{\text{released}} = -230 \text{ kJ} \][/tex]
Using the moles of Fe₂O₃ per kJ calculated in Step 1:
[tex]\[ \text{Moles of Fe₂O₃ formed} = -0.001176470588235294 \text{ moles per kJ} \times -230 \text{ kJ} = 0.27058823529411763 \text{ moles} \][/tex]
### Step 3: Calculate the molar mass of Fe₂O₃
The molar mass of Fe₂O₃ is calculated by summing the atomic masses of its constituent elements:
[tex]\[ \text{Molar mass of Fe₂O₃} = 2 \times 55.845 \text{ g/mol} + 3 \times 16.00 \text{ g/mol} = 159.69 \text{ g/mol} \][/tex]
### Step 4: Calculate the grams of Fe₂O₃ formed
Using the moles of Fe₂O₃ formed from Step 2 and the molar mass from Step 3:
[tex]\[ \text{Grams of Fe₂O₃ formed} = 0.27058823529411763 \text{ moles} \times 159.69 \text{ g/mol} = 43.210235294117645 \text{ grams} \][/tex]
Considering we need to express the answer using two significant figures:
[tex]\[ \text{Grams of Fe₂O₃ formed} \approx 43 \text{ grams} \][/tex]
### Answer
The amount of rust (Fe₂O₃) formed when 230 kJ are released is approximately 43 grams.
