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The problem asks to determine the mass of FeSO₄ · 7H₂O in a sample of 3.660 g based on the mass of Fe₂O₃ (0.472 g) obtained after a series of reactions.

Let's break down the solution step by step:

### 1. Chemical Reaction Involved
When FeSO₄ · 7H₂O is dissolved and treated, iron (Fe²⁺) is converted to Fe³⁺, which precipitates as Fe₂O₃ after heating. The key is to relate the amount of Fe₂O₃ produced to the original amount of FeSO₄ · 7H₂O.

- **Molar mass of Fe₂O₃**: 
  \[
  \text{Fe}_2\text{O}_3 = 2 \times 55.85 \, (\text{Fe}) + 3 \times 16 \, (\text{O}) = 159.7 \, \text{g/mol}
  \]

- **Moles of Fe₂O₃ produced**:
  \[
  \text{moles of Fe}_2\text{O}_3 = \frac{\text{mass of Fe}_2\text{O}_3}{\text{molar mass of Fe}_2\text{O}_3} = \frac{0.472 \, \text{g}}{159.7 \, \text{g/mol}} = 0.002955 \, \text{mol}
  \]

### 2. Determine Moles of Iron (Fe³⁺)
Fe₂O₃ contains two moles of Fe atoms for every mole of Fe₂O₃.

- **Moles of Fe** in 0.472 g of Fe₂O₃:
  \[
  \text{moles of Fe} = 2 \times 0.002955 \, \text{mol} = 0.00591 \, \text{mol}
  \]

### 3. Relate to FeSO₄ · 7H₂O
Each mole of Fe in FeSO₄ · 7H₂O corresponds to one mole of FeSO₄ · 7H₂O.

- **Moles of FeSO₄ · 7H₂O**:
  \[
  \text{moles of FeSO₄ · 7H₂O} = 0.00591 \, \text{mol}
  \]

### 4. Calculate the Mass of FeSO₄ · 7H₂O
Now, using the molar mass of FeSO₄ · 7H₂O (278.01 g/mol):

- **Mass of FeSO₄ · 7H₂O**:
  \[
  \text{mass of FeSO₄ · 7H₂O} = 0.00591 \, \text{mol} \times 278.01 \, \text{g/mol} = 1.643 \, \text{g}
  \]

Thus, the mass of FeSO₄ · 7H₂O in the 3.660 g sample is **1.643 g**.

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Would you like further details on any part of the calculation? Here are some additional questions to explore this problem further:

1. How does the mass of FeSO₄ · 7H₂O compare to the total sample mass of 3.660 g?
2. What assumptions are made when calculating the moles of iron from Fe₂O₃?
3. How would the result change if the sample had been impure or contained other iron salts?
4. Can you describe the reaction steps converting Fe²⁺ to Fe₂O₃?
5. Why is NH₃ added to the reaction to precipitate Fe₂O₃?

**Tip**: Always double-check stoichiometric ratios when converting between compounds, especially in precipitation reactions!

A powder contains FeSO₄ · 7H₂O (molar mass = 278.01 g/mol), among other components. A 3.660 g sample of the powder was dissolved in HNO₃ and heated to convert all iron to Fe³⁺. The addition of NH₃ precipitated Fe₂O₃ · xH₂O, which was subsequently ignited to produce 0.472 g of Fe₂O₃. What was the mass of FeSO₄ · 7H₂O in the 3.660 g sample?

Learn how to calculate the mass of FeSO₄ · 7H₂O in a 3.660 g sample after reacting with HNO₃ and precipitating Fe₂O₃. This step-by-step solution covers stoichiometry, the mole concept, and mass calculation. Ideal for Grades 11-12 chemistry students.

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