| ⇦ | 
| ⇨ |
How many FeSO₄.7H₂SO₄ will be oxidised by an acidified solution containing 9.48 g KMnO₄
Options
(a) 83.4 g
(b) 16.7 g
(c) 1.67 g
(d) 8.3 g
Correct Answer:
83.4 g
Explanation:
Balanced reaction: 2KMnO₄ + 10FeSO₄.7H₂O + 8H₂SO₄ → 5F₂(SO₄)₃ + K₂SO₄ + 2MnSO₄ + 78H₂O. 2 * 158 g KMnO₄ oxidises = 10 * 278 g FeSO₄.7H₂O. 9.48 g KMnO₄ oxidises = 10 * 278 * 9.48 / 2 * 158 = 83.4 g.
Related Questions: - In the cyanide ion, the formal negative charge is on
- How many moles of iodine are liberated when 1 mole of potassium dichromate
- The oxidation number of P in Mg₂P₂O₇ is
- Which of the following compound is not coloured
- Which of the following statements is not true about alcohols
Topics: D and F Block Elements
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the cyanide ion, the formal negative charge is on
- How many moles of iodine are liberated when 1 mole of potassium dichromate
- The oxidation number of P in Mg₂P₂O₇ is
- Which of the following compound is not coloured
- Which of the following statements is not true about alcohols
Topics: D and F Block Elements (91)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply