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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: - In the periodic table going down in fluorine group
- In Cr₂O₇²⁻ the oxidation numbers are
- Which compound is electron deficient?
- Which of the following processes does not involve oxidation of iron
- Oxidation number of iron in [Fe(CN)₆]⁴⁻ is
Topics: D and F Block Elements
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the periodic table going down in fluorine group
- In Cr₂O₇²⁻ the oxidation numbers are
- Which compound is electron deficient?
- Which of the following processes does not involve oxidation of iron
- Oxidation number of iron in [Fe(CN)₆]⁴⁻ is
Topics: D and F Block Elements (91)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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