To determine the number of moles of iron(II) hydroxide [tex]\( \text{Fe(OH)}_2 \)[/tex] produced from the reaction
[tex]\[
\text{FeCl}_2 + 2 \text{KOH} \rightarrow \text{Fe(OH)}_2 + 2 \text{KCl},
\][/tex]
we need to identify the limiting reagent. A limiting reagent is the substance that is entirely consumed first in a chemical reaction and thus determines the maximum amount of product that can be formed.
1. Calculate the mole ratio for each reactant:
- The balanced equation shows that 1 mole of [tex]\( \text{FeCl}_2 \)[/tex] reacts with 2 moles of [tex]\( \text{KOH} \)[/tex].
- Given:
- [tex]\( \text{FeCl}_2 \)[/tex] = 4.15 moles
- [tex]\( \text{KOH} \)[/tex] = 3.62 moles
2. Determine the ratios:
- From the balanced chemical equation, we need 1 mole of [tex]\( \text{FeCl}_2 \)[/tex] for every 2 moles of [tex]\( \text{KOH} \)[/tex].
- Calculate the effective ratios:
- For [tex]\( \text{FeCl}_2 \)[/tex]: [tex]\( \boxed{\frac{4.15}{1}} = 4.15 \)[/tex]
- For [tex]\( \text{KOH} \)[/tex]: [tex]\( \boxed{\frac{3.62}{2}} = 1.81 \)[/tex]
3. Identify the limiting reagent:
- The limiting reagent is the reactant with the smaller ratio. Here, [tex]\( 1.81 \)[/tex] (from [tex]\( \text{KOH} \)[/tex]) is smaller than [tex]\( 4.15 \)[/tex] (from [tex]\( \text{FeCl}_2 \)[/tex]).
4. Determine the moles of [tex]\( \text{Fe(OH)}_2 \)[/tex] produced:
- Since [tex]\( \text{KOH} \)[/tex] is the limiting reagent:
- Each mole of [tex]\( \text{KOH} \)[/tex] can be used to form [tex]\( \boxed{\frac{1}{2}} \)[/tex] mole of [tex]\( \text{Fe(OH)}_2 \)[/tex].
- Since the limiting reagent amount is 1.81 moles, it will produce:
- [tex]\( \boxed{1.81} \)[/tex] moles of [tex]\( \text{Fe(OH)}_2 \)[/tex].
So, the reaction will produce 1.81 moles of iron(II) hydroxide. Therefore, the correct answer is:
A. [tex]\( 1.81 \)[/tex] mol
