In the reaction between 0.060 mol of aluminium metal and iron(ll) oxide, 2.000 g of iron is produced. How many mol of iron (ll) oxide reacted? 2Al() Fe203(s) = 2Fe() A1203(s)



Answer :

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 \( \boxed{0.060 \text{ mol}} \).