| ⇦ | 
| ⇨ |
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: - The C – C bond length in ethane is
- Phenolic carboxylic acids may be prepared by
- The bond length in LiF will be
- Which of the following compounds gives trichloromethane on distilling with
- The hybridisation of B in BF₃ molecules is
Topics: D and F Block Elements
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The C – C bond length in ethane is
- Phenolic carboxylic acids may be prepared by
- The bond length in LiF will be
- Which of the following compounds gives trichloromethane on distilling with
- The hybridisation of B in BF₃ molecules is
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