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A chemist is working on a reaction represented by this chemical equation:

[tex]\[ FeCl_2 + 2 KOH \rightarrow Fe(OH)_2 + 2 KCl \][/tex]

If the chemist uses 4.15 moles of iron(II) chloride and 3.62 moles of potassium hydroxide, how many moles of iron(II) hydroxide will the reaction produce?

A. [tex]\(\quad 1.81 \, \text{mol}\)[/tex]

B. [tex]\(\quad 3.62 \, \text{mol}\)[/tex]

C. [tex]\(\quad 4.15 \, \text{mol}\)[/tex]

D. [tex]\(\quad 7.24 \, \text{mol}\)[/tex]

Sagot :

GinnyAnsweridn
Let's solve this problem step by step by taking into account the given amounts of reactants and comparing them to the stoichiometry of the reaction.

1. Reaction Analysis:
- The balanced chemical equation for the reaction is:
[tex]\[ FeCl_2 + 2 KOH \rightarrow Fe(OH)_2 + 2 KCl \][/tex]
This tells us 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 Limiting Reactant:
- We need to compare the mole ratio of the reactants provided to find out which one will limit the reaction.

3. Iron(II) Chloride ([tex]\( \text{FeCl}_2 \)[/tex]):
- The chemist uses 4.15 moles of [tex]\( \text{FeCl}_2 \)[/tex].
- According to the balanced equation, 4.15 moles of [tex]\( \text{FeCl}_2 \)[/tex] would ideally produce 4.15 moles of [tex]\( \text{Fe(OH)}_2 \)[/tex] (since the ratio is 1:1). This is because each mole of [tex]\( \text{FeCl}_2 \)[/tex] produces 1 mole of [tex]\( \text{Fe(OH)}_2 \)[/tex].

4. Potassium Hydroxide ([tex]\( \text{KOH} \)[/tex]):
- The chemist uses 3.62 moles of [tex]\( \text{KOH} \)[/tex].
- According to the balanced equation, 2 moles of [tex]\( \text{KOH} \)[/tex] are required to react with 1 mole of [tex]\( \text{FeCl}_2 \)[/tex]. Therefore, we need to find out how much [tex]\( \text{Fe(OH)}_2 \)[/tex] can be produced from 3.62 moles of [tex]\( \text{KOH} \)[/tex]:
[tex]\[ \frac{3.62 \, \text{moles} \, \text{KOH}}{2 \, \text{moles} \, \text{KOH} / \text{mole} \, \text{Fe(OH)}_2} = 1.81 \, \text{moles} \, \text{Fe(OH)}_2 \][/tex]

5. Comparison and Determination of Limiting Reactant:
- [tex]\( \text{FeCl}_2 \)[/tex] can theoretically produce 4.15 moles of [tex]\( \text{Fe(OH)}_2 \)[/tex].
- [tex]\( \text{KOH} \)[/tex] can produce only 1.81 moles of [tex]\( \text{Fe(OH)}_2 \)[/tex].

The limiting reactant is [tex]\( \text{KOH} \)[/tex] because it produces the least amount of [tex]\( \text{Fe(OH)}_2 \)[/tex].

6. Final Amount of [tex]\( \text{Fe(OH)}_2 \)[/tex] Produced:
- Since [tex]\( \text{KOH} \)[/tex] is the limiting reactant, it determines the amount of [tex]\( \text{Fe(OH)}_2 \)[/tex] produced. Therefore, the reaction will produce 1.81 moles of [tex]\( \text{Fe(OH)}_2 \)[/tex].

Thus, the correct answer is:

A. [tex]\( 1.81 \)[/tex] mol
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