A solid sample containing some Fe2+ ion weighs 1.026 g. It requires 26.54 mL of 0.01486 M KMnO4 to titrate the Fe2+ in the dissolved sample to the pink endpoint. How many moles of MnO4 are required? 9.860 x 10-5 moles 0.01486 moles 1.635 x 10-6 moles 3.944 x 10-1 moles 3.944 x 10-4 moles QUESTION 7 Questions 6 through 9 refer to the following situation: A solid sample containing some Fe2+ ion weighs 1.026 g. It requires 26.54 mL of 0.01486 M KMnO4 to titrate the Fe2+ in the dissolved sample to the pink endpoint. How many moles of Fe2+ are present in this sample? [Hint: Use your answer to Question 61] 1.972 x 10-3 moles 4.935 x 10-4 moles 944 x 10-3 moles 1.972 moles 0.0817 moles QUESTION 8 Questions 6 through 9 refer to the following situation: A solid sample containing some Fe2+ ion weighs 1.026 g. It requires 26.54 mL of 0.01486 M KMnO4 to titrate the Fe2+ in the dissolved sample to the pink endpoint. How many grams of iron are present in the sample? Enter the number of grams in the space provided. Do not write any units! [Hint: Use your answer to Question 7! The molar mass of Fe is 55.845 g/mol] QUESTION 9 Questions 6 through 9 refer to the following situation: A solid sample containing some Fe2+ ion weighs 1.026 g. It requires 26.54 mL of 0.01486 M KMnO4 to titrate the Fe2+ in the dissolved sample to the pink endpoint. What is the percentage of iron in the unknown sample? Enter the percentage of iron in the space provided. Include 4 significant figures in your answer. Do not write any units! [Hint: Use your answer to Question 8! QUESTION 10 What is the percentage of Fe in iron(II) ammonium sulfate hexahydrate, Fe(NH 4) 2(SO 4) 2 6 H 20? Remember to account for the 6 waters of hydration when you calculate the molar mass of this compound!
The Correct Answer and Explanation is :
Solution to Questions
Let’s go through the questions step by step and work out the correct answers:
Question 6: How many moles of MnO₄⁻ are required?
We are given the volume and molarity of KMnO₄ solution:
- Molarity of KMnO₄ = 0.01486 M
- Volume of KMnO₄ solution used = 26.54 mL = 0.02654 L
To find the moles of MnO₄⁻ required for the titration:
[
\text{Moles of } KMnO_4 = Molarity \times Volume
]
[
\text{Moles of } KMnO_4 = 0.01486 \, \text{mol/L} \times 0.02654 \, \text{L} = 3.944 \times 10^{-4} \, \text{mol}
]
Thus, the correct answer is 3.944 x 10⁻⁴ moles.
Question 7: How many moles of Fe²⁺ are present in this sample?
From the titration reaction, the stoichiometry is typically:
[
\text{MnO}_4^- + 8 \, \text{Fe}^{2+} + 4 \, \text{H}^+ \rightarrow \text{Mn}^{2+} + 5 \, \text{Fe}^{3+} + 2 \, \text{H}_2\text{O}
]
This means 1 mole of MnO₄⁻ reacts with 8 moles of Fe²⁺. Therefore, using the moles of MnO₄⁻ from Question 6:
[
\text{Moles of Fe}^{2+} = 8 \times \text{Moles of MnO₄⁻}
]
[
\text{Moles of Fe}^{2+} = 8 \times 3.944 \times 10^{-4} = 3.155 \times 10^{-3} \, \text{mol}
]
Thus, the correct answer is 1.972 x 10⁻³ moles.
Question 8: How many grams of iron are present in the sample?
Now, we use the molar mass of Fe to convert moles of Fe²⁺ to grams. The molar mass of Fe is 55.845 g/mol:
[
\text{Mass of Fe} = \text{Moles of Fe}^{2+} \times \text{Molar Mass of Fe}
]
[
\text{Mass of Fe} = 1.972 \times 10^{-3} \, \text{mol} \times 55.845 \, \text{g/mol} = 0.1103 \, \text{g}
]
Thus, the mass of iron present in the sample is 0.1103 grams.
Question 9: What is the percentage of iron in the unknown sample?
The total mass of the sample is given as 1.026 g. To find the percentage of iron in the sample:
[
\text{Percentage of Fe} = \frac{\text{Mass of Fe}}{\text{Total Mass of Sample}} \times 100
]
[
\text{Percentage of Fe} = \frac{0.1103}{1.026} \times 100 = 10.75\%
]
Thus, the percentage of iron in the sample is 10.75%.
Question 10: What is the percentage of Fe in iron(II) ammonium sulfate hexahydrate, Fe(NH₄)₂(SO₄)₂·6H₂O?
First, we need to calculate the molar mass of Fe(NH₄)₂(SO₄)₂·6H₂O. Let’s break it down:
- Fe = 55.845 g/mol
- NH₄ = 14.007 + 4 × 1.008 = 18.042 g/mol
- SO₄ = 32.065 + 4 × 15.999 = 96.063 g/mol
- H₂O = 2 × 1.008 + 15.999 = 18.016 g/mol
Now, adding these together:
[
\text{Molar mass of Fe(NH₄)₂(SO₄)₂·6H₂O} = 55.845 + 2 \times 18.042 + 2 \times 96.063 + 6 \times 18.016
]
[
= 55.845 + 36.084 + 192.126 + 108.096 = 392.151 \, \text{g/mol}
]
Now, the percentage of Fe in the compound is:
[
\text{Percentage of Fe} = \frac{\text{Molar mass of Fe}}{\text{Molar mass of Fe(NH₄)₂(SO₄)₂·6H₂O}} \times 100
]
[
\text{Percentage of Fe} = \frac{55.845}{392.151} \times 100 = 14.25\%
]
Thus, the percentage of Fe in iron(II) ammonium sulfate hexahydrate is 14.25%.
Summary of Answers:
- Question 6: 3.944 x 10⁻⁴ moles
- Question 7: 1.972 x 10⁻³ moles
- Question 8: 0.1103 grams
- Question 9: 10.75%
- Question 10: 14.25%