मराठी

When heated, potassium permanganate decomposes according to the following equation: 2KMnOA4⟶KA2MnOA4+MnOA2solid residue+OA2 Given that the molecular mass of potassium permanganate is 158 g,

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प्रश्न

When heated, potassium permanganate decomposes according to the following equation:

\[\ce{2KMnO4 -> \underset{solid residue}{K2MnO4 + MnO2} + O2}\]

Given that the molecular mass of potassium permanganate is 158 g, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres). [K = 39, Mn = 55, O = 16]

संख्यात्मक
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उत्तर

2KMnO4 →  K2MnO4 + MnO2 + O2
2[39 + 55 + 64]           24 lit
2 × 158 = 316 g            

Vol. of O2 produced by 316 g of KMnO= 24 lit.

∴ Vol. of O2 produced by 15.8 of KMnO4 = `(24 xx 15.8)/316`

= 1.2 lit.

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Mole Concept and Stoichiometry - Exercise 6 [पृष्ठ १२०]

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फ्रँक Chemistry Part 2 [English] Class 10 ICSE
पाठ 5 Mole Concept and Stoichiometry
Exercise 6 | Q 3.2 | पृष्ठ १२०
एस.पी. सिंह Concise Chemistry [English] Class 10 ICSE
पाठ 5 Mole concept and Stoichiometry
MISCELLANEOUS EXERCISE | Q 39. (b) | पृष्ठ ९८

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