| ⇦ | 
| ⇨ |
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: - CH₃CH₂OH converts into CH₃CHO in the presence of
- Petroleum is a mixture of
- In Ramsay and Rayleigh’s isolation on noble gases from air,the nitrogen of the
- The basic component of photochemical smog is
- A molal solution is one that contains one mole of a solute in
Topics: D and F Block Elements
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- CH₃CH₂OH converts into CH₃CHO in the presence of
- Petroleum is a mixture of
- In Ramsay and Rayleigh’s isolation on noble gases from air,the nitrogen of the
- The basic component of photochemical smog is
- A molal solution is one that contains one mole of a solute in
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