Let's solve this problem step-by-step.
1. Identify the reaction and the stoichiometric coefficients:
The balanced chemical equation for the reaction given is:
[tex]\[
FeCl_2 + 2 KOH \rightarrow Fe(OH)_2 + 2 KCl
\][/tex]
From this equation, we see that 1 mole of [tex]\( FeCl_2 \)[/tex] reacts with 2 moles of [tex]\( KOH \)[/tex] to produce 1 mole of [tex]\( Fe(OH)_2 \)[/tex].
2. Determine the moles of reagents:
- Moles of [tex]\( FeCl_2 \)[/tex]: 4.15
- Moles of [tex]\( KOH \)[/tex]: 3.62
3. Calculate the required moles of [tex]\( KOH \)[/tex] for 4.15 moles of [tex]\( FeCl_2 \)[/tex]:
According to the stoichiometric ratio, 1 mole of [tex]\( FeCl_2 \)[/tex] requires 2 moles of [tex]\( KOH \)[/tex]. Therefore, 4.15 moles of [tex]\( FeCl_2 \)[/tex] would require:
[tex]\[
4.15 \text{ moles of } FeCl_2 \times 2 \text{ moles of } KOH / 1 \text{ mole of } FeCl_2 = 8.3 \text{ moles of } KOH
\][/tex]
4. Compare the available moles of [tex]\( KOH \)[/tex] with the required moles:
- Required moles of [tex]\( KOH \)[/tex]: 8.3
- Available moles of [tex]\( KOH \)[/tex]: 3.62
Since the available moles of [tex]\( KOH \)[/tex] (3.62) are less than the required moles (8.3), [tex]\( KOH \)[/tex] is the limiting reagent.
5. Determine the amount of product ([tex]\( Fe(OH)_2 \)[/tex]) produced:
Since [tex]\( KOH \)[/tex] is the limiting reagent, we need to find out how many moles of [tex]\( Fe(OH)_2 \)[/tex] can be produced by the available [tex]\( KOH \)[/tex].
According to the stoichiometric ratio, 2 moles of [tex]\( KOH \)[/tex] produce 1 mole of [tex]\( Fe(OH)_2 \)[/tex]. Hence, 3.62 moles of [tex]\( KOH \)[/tex] will produce:
[tex]\[
\frac{3.62 \text{ moles of } KOH}{2} = 1.81 \text{ moles of } Fe(OH)_2
\][/tex]
Therefore, the reaction will produce 1.81 moles of iron(II) hydroxide.
The correct answer is:
A. [tex]$1.81 \text{ mol}$[/tex]
