Question

A sample of KClO₃ on decomposition yeilded 448 mL of oxygen gas at NTP. Calculate : 

(i) weight of oxygen produced , 

(ii) weight of KClO₃ originaly taken 

(iii) weight of KCl produced ( K = 39 , Cl = 35.5 and O = 16 ) 

Answer 1

(i) Mole of oxygen = 448 / 22400 = 0.02 

Wt. of oxygen = 0.02 x 32 = 0.64 gm

Applying POAC on O atoms 

Moles of O atoms in KClO₃ = moles of O atoms in O₂ 

3 (moles of KClO₃) = 2 (moles of O₂) 

(1 mole of KClO₃ contains 3 moles of O and 1 mole of O₂ contains 2 moles of O) 

Wt. of KClO3 = 1.634 g 

(iii) Again applying POAC for K atoms, 

Moles of K atoms in KClO₃ = 1 x moles of KCl 

(1 mole of KClO₃ contains 1 mole of K and 1 mole of KCl contains 1 mole of K) 

Wt. of KCl = 0.9937 g.