At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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
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
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.