To determine the number of moles of iron(II) hydroxide [tex]\(\text{Fe(OH)}_2\)[/tex] produced in the reaction, we need to follow these steps:
1. Write down the balanced chemical equation:
[tex]\[
\text{FeCl}_2 + 2 \text{KOH} \rightarrow \text{Fe(OH)}_2 + 2\text{KCl}
\][/tex]
This equation shows that 1 mole of [tex]\(\text{FeCl}_2\)[/tex] reacts with 2 moles of [tex]\(\text{KOH}\)[/tex] to produce 1 mole of [tex]\(\text{Fe(OH)}_2\)[/tex].
2. Determine the molar ratio:
According to the balanced equation:
- 1 mole of [tex]\(\text{FeCl}_2\)[/tex] produces 1 mole of [tex]\(\text{Fe(OH)}_2\)[/tex]
- 2 moles of [tex]\(\text{KOH}\)[/tex] react with 1 mole of [tex]\(\text{FeCl}_2\)[/tex]
3. Determine the limiting reactant:
- Given [tex]\(\text{4.15 moles of FeCl}_2\)[/tex]
- Given [tex]\(\text{3.62 moles of KOH}\)[/tex]
To find the limiting reactant, compare the mole ratios:
Calculate the required amount of [tex]\(\text{FeCl}_2\)[/tex] to react completely with [tex]\(\text{KOH}\)[/tex]:
- Moles of [tex]\(\text{FeCl}_2\)[/tex] needed = [tex]\(\frac{3.62 \text{ moles of KOH}}{2 \text{ moles of KOH/mole of FeCl}_2}\)[/tex]
- Moles of [tex]\(\text{FeCl}_2\)[/tex] needed = [tex]\(\frac{3.62}{2}\)[/tex]
- Moles of [tex]\(\text{FeCl}_2\)[/tex] needed = 1.81
Since we have 4.15 moles of [tex]\(\text{FeCl}_2\)[/tex] available and only need 1.81 moles to react with the 3.62 moles of [tex]\(\text{KOH}\)[/tex], [tex]\(\text{KOH}\)[/tex] is the limiting reactant.
4. Calculate the moles of [tex]\(\text{Fe(OH)}_2\)[/tex] produced:
- Since [tex]\(\text{KOH}\)[/tex] is the limiting reactant and every 2 moles of [tex]\(\text{KOH}\)[/tex] produce 1 mole of [tex]\(\text{Fe(OH)}_2\)[/tex]:
- Moles of [tex]\(\text{Fe(OH)}_2\)[/tex] produced = [tex]\(\frac{3.62 \text{ moles of KOH}}{2 \text{ moles of KOH/mole of Fe(OH}}_2)\)[/tex]
- Moles of [tex]\(\text{Fe(OH)}_2\)[/tex] produced = [tex]\(\frac{3.62}{2}\)[/tex]
- Moles of [tex]\(\text{Fe(OH)}_2\)[/tex] produced = 1.81
Therefore, the reaction will produce 1.81 moles of iron(II) hydroxide [tex]\(\text{Fe(OH)}_2\)[/tex].
The correct answer is:
A. 1.81 mol
