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)

Answer 1

Molar mass of KNO₃ = [39 + 14 + (3 $\times$ 16)] g = 101 g mol⁻¹
Molar mass of KNO₂ = [39 + 14 + (2 $\times$ 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 $\times$ Molecular weight of KNO₂ = $\frac{5.05}{101}\times 85=4.25\mathrm{g}$

(ii) Weight of oxygen formed = Number of moles $\times$ Molecular weight of oxygen = $\frac{1}{2}\left(\frac{5.05}{101}\right)\times 32=0.8\mathrm{g}$