What is the % Iron in an FeSO4 Sample Based on Titration Data?

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Discussion Overview

The discussion revolves around determining the percentage of iron in an impure FeSO4 sample based on titration data involving KMnO4. Participants explore the chemical reactions involved, the calculations necessary to derive the percentage, and the nature of the titration reaction.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Post 1 presents a titration scenario and initial calculations to find the grams of FeSO4, seeking guidance on how to derive grams of Fe from that value.
  • Post 2 questions the nature of the reaction, suggesting it may not be an oxidation-reduction reaction as initially stated.
  • Post 3 proposes the possibility of the reaction being an acid-base reaction, referencing a general reaction format.
  • Post 4 challenges the assumptions about the nature of KMnO4 and FeSO4, emphasizing the expectation of an oxidation-reduction reaction and requesting balanced half-reactions.
  • Post 5 provides a proposed balanced reaction involving KMnO4 and FeSO4, but it is unclear if this is accepted as correct.
  • Post 6 offers a detailed calculation for determining the endpoint and the resulting grams of Fe, concluding with a percentage of iron, but does not confirm the accuracy of the reaction balance.
  • Post 7 critiques the balanced reaction provided in Post 6, claiming it is not balanced, indicating a potential error in the calculations or assumptions made.

Areas of Agreement / Disagreement

Participants express disagreement regarding the nature of the reaction (oxidation-reduction vs. acid-base) and the correctness of the balanced equations. There is no consensus on the balanced reaction or the calculations leading to the percentage of iron.

Contextual Notes

Participants highlight potential issues with the balanced reaction and the calculations, indicating that assumptions about the reaction type and stoichiometry may affect the results. The discussion remains focused on refining these aspects without reaching a definitive conclusion.

Who May Find This Useful

This discussion may be useful for students studying analytical chemistry, particularly those interested in titration techniques and reaction stoichiometry.

sami23
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Homework Statement



A 0.6500 g sample of an impure FeSO4 sample is titrated with 33.25 mL of 0.0190 M KMnO4 to an endpoint. What is the % of iron in the sample?


Homework Equations



KMnO4 + FeSO4 --> KSO4 + FeMnO4

% Fe = (grams of Fe / grams of sample) * 100%

The Attempt at a Solution



0.03325 L sol'n * (0.0190 M KMnO4 / 1 L sol'n) = 6.3175*10-4 M KMnO4

g FeSO4 = 6.3175*10-4 M KMnO4 * (1 mol FeSO4 / 1 mol KMnO4) * (151.903 g FeSO4 / 1 mol FeSO4 = 0.09596 g FeSO4

But now I'm stuck. How do I find g of Fe from 0.09596 g FeSO4?

% Fe = (grams Fe alone/ 0.6500g FeSO4)*100% = final answer.
 
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What kind of reaction occurs in the analytical titration? Is this an oxidation-reduction reaction? If it is, your reaction statement is wrong.
 
Does that mean it's an acid-base reaction. If it is, then acid + base --> salt + water
 
Sami233, how do you come to the original question that you asked? Are you enrolled in a relevant course now for which you are trying to answer the exercise question?

KMnO4 is not necessarily an acid or base. FeSO4 is not necessarily an acid or base. What do you expect to occur in the reaction between the two? I strongly expect the reaction to be an oxidation-reduction one. Can you write a balanced reaction for this? Can you start with the two separate half-reactions? What kind of endpoint would you expect to observe?
 
2 KMnO4 + FeSO4 --> Fe(MnO4)2 + K2SO4
Fe + H2SO4 --> FeSO4 + H2
 
Endpoint can be determined by:
2MnO4 + 10FeSO4 + 8H2SO4 --> 5Fe(SO4)3 + 2MnSO4 + K2SO4 + 8H2O

0.03325 L * (0.019 M KMnO4/1L) = 6.3175 x 10-4 mol KMnO4

6.3175 x 10-4 mol KMnO4 * (10 mol FeSO4/2 mol MnSO4) * (151.9 g FeSO4/1 mol FeSO4) = 0.4798 g FeSO4

0.4798 g FeSO4 * (1 mol FeSO4/151.9 g FeSO4)* (55.847 g Fe/1 mole Fe) = 0.1764 g Fe

% Fe = (0.1764 g Fe/0.6500 g sample)*100% = 27.139%
 
Endpoint can be determined by:
2MnO4 + 10FeSO4 + 8H2SO4 --> 5Fe(SO4)3 + 2MnSO4 + K2SO4 + 8H2O

That reaction is not balanced.
 

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