| ⇦ |
| ⇨ |
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 gas phase reaction C₂H₂ + H₂ ⇌ C₂H₆, the equilibrium constant
- Pyridine is less basic than triethylamine because
- Decreasing order of nucleophilicity is
- What products are formed when the following compound is treated with Br₂
- A compound with empirical formula CH₂O has a vapour density of 30.Its molecular
Topics: D and F Block Elements
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the gas phase reaction C₂H₂ + H₂ ⇌ C₂H₆, the equilibrium constant
- Pyridine is less basic than triethylamine because
- Decreasing order of nucleophilicity is
- What products are formed when the following compound is treated with Br₂
- A compound with empirical formula CH₂O has a vapour density of 30.Its molecular
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