| ⇦ | 
| ⇨ |
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: - Boron has two stable isotopes, ¹°B(19%) and ¹¹B(81%). Average atomic weight
- p-Nitrobromobenzene can be converted to p-nitroaniline by using NaNH₂.
- The wrong statement about fullerene is
- Decreasing order of stability of O₂ , O₂⁻ , O₂⁺ and O₂²⁻ is
- K₂Cr₂O₇ on heating with aqueous NaOH gives
Topics: D and F Block Elements
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Boron has two stable isotopes, ¹°B(19%) and ¹¹B(81%). Average atomic weight
- p-Nitrobromobenzene can be converted to p-nitroaniline by using NaNH₂.
- The wrong statement about fullerene is
- Decreasing order of stability of O₂ , O₂⁻ , O₂⁺ and O₂²⁻ is
- K₂Cr₂O₇ on heating with aqueous NaOH gives
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