| ⇦ |
| ⇨ |
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: - Standard enthalpy of vaporisation Δvap H⁰ for water at 100⁰C is 40.66 kJ mol⁻¹.
- Number of isomers of C₄H₁₀ is
- When sodium benzoate reacts with sodalime what is produced
- Choose isosteres from the following.
- Which of the following is a pair of isoelectronic species?
Topics: D and F Block Elements
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Standard enthalpy of vaporisation Δvap H⁰ for water at 100⁰C is 40.66 kJ mol⁻¹.
- Number of isomers of C₄H₁₀ is
- When sodium benzoate reacts with sodalime what is produced
- Choose isosteres from the following.
- Which of the following is a pair of isoelectronic species?
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