| ⇦ | 
| ⇨ |
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: - When aniline is treated with chloroform in the presence of alcoholic KOH, the product
- If the energy of a photon is given as : =3.03 × 10⁻¹⁹J then, the wavelength (λ)
- Chromium has the most stable oxidation state
- The oxidation state of sulpur in sodium tetrathionate (Na₂S₄O₆) is
- Chlorine is in +1 oxidation state in which of the following
Topics: D and F Block Elements
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When aniline is treated with chloroform in the presence of alcoholic KOH, the product
- If the energy of a photon is given as : =3.03 × 10⁻¹⁹J then, the wavelength (λ)
- Chromium has the most stable oxidation state
- The oxidation state of sulpur in sodium tetrathionate (Na₂S₄O₆) is
- Chlorine is in +1 oxidation state in which of the following
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