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A chemist is working on a reaction represented by the following 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. 1.81 mol
B. 3.62 mol
C. 4.15 mol
D. 7.24 mol

Sagot :

GinnyAnswer
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
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