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Question

Potassium nitrate on strong heating decomposes as under:
2KNO3 2KNO2 + O2
Calculate:
(i) weight of potassium nitrite formed.
(ii) weight of oxygen formed when 5.05 g of potassium nitrate decomposes completely.
(K = 39, O = 16, N = 14)

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Solution

Molar mass of KNO3 = [39 + 14 + (3 × 16)] g = 101 g mol1
Molar mass of KNO2 = [39 + 14 + (2 × 16)] g = 85 g mol1
According to balanced chemical reaction, two moles of KNO3 decomposes to form two moles of KNO2 and one mole of oxygen gas.
Number of moles of KNO3 decomposed = Number of moles of KNO2 formed = 12Number of moles of O2 formed
It is given that mass of KNO2 taken for the reaction is 5.054 grams.

(i) Weight of potassium nitrite formed = Number of moles × Molecular weight of KNO2 = 5.05101×85=4.25 g

(ii) Weight of oxygen formed = Number of moles × Molecular weight of oxygen = 125.05101×32=0.8 g


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