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Question

# Potassium nitrate on strong heating decomposes as under: 2KNO3 $\stackrel{∆}{\to }$ 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 mol−1 Molar mass of KNO2 = [39 + 14 + (2 $×$ 16)] g = 85 g mol−1 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 = $\frac{1}{2}$Number 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 = $\frac{5.05}{101}×85=4.25\mathrm{g}$ (ii) Weight of oxygen formed = Number of moles $×$ Molecular weight of oxygen = $\frac{1}{2}\left(\frac{5.05}{101}\right)×32=0.8\mathrm{g}$

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