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Sagot :
Answer:
Explanation:
To determine how many moles of iron(II) oxide (\(Fe_2O_3\)) reacted, we need to follow these steps:
1. **Calculate the moles of iron produced:**
Given that 0.060 mol of aluminum reacts and produces 2.000 g of iron, we first find the molar mass of iron to convert grams to moles.
Molar mass of iron (\(Fe\)) = 55.845 g/mol
Number of moles of iron produced:
\[
\text{moles of } Fe = \frac{\text{mass of } Fe}{\text{molar mass of } Fe} = \frac{2.000 \text{ g}}{55.845 \text{ g/mol}} = 0.0358 \text{ mol}
\]
2. **Apply the stoichiometry of the reaction:**
From the balanced chemical equation:
\[
2Al + Fe_2O_3 \rightarrow 2Fe + Al_2O_3
\]
The stoichiometric ratio between aluminum and iron is 2:2 (or 1:1), which means for every mole of aluminum that reacts, one mole of iron(II) oxide reacts.
3. **Calculate moles of \(Fe_2O_3\) reacted:**
Since 0.060 mol of aluminum reacted, the same number of moles of \(Fe_2O_3\) must have reacted:
\[
\text{moles of } Fe_2O_3 = 0.060 \text{ mol}
\]
Therefore, the number of moles of iron(II) oxide (\(Fe_2O_3\)) that reacted is \( \boxed{0.060 \text{ mol}} \).
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