A 12.25 g sample of pure KClO 3 (Molar mass 122.5 g/mol ) is heated. It decomposes via two parallel pathways: (I) 2 KClO 3 ( s ) → 2 KCl ( s ) + 3 O 2 ( g ) (II) 4 KClO 3 ( s ) → 3 KClO 4 (

Question

A 12.25 g sample of pure KClO 3 ​ (Molar mass 122.5 g/mol ) is heated. It decomposes via two parallel pathways: (I) 2 KClO 3 ​ ( s ) → 2 KCl ( s ) + 3 O 2 ​ ( g ) (II) 4 KClO 3 ​ ( s ) → 3 KClO 4 ​ ( s ) + KCl ( s ) After heating, the total mass of solid residue obtained is 8.54 g . What is the mass percentage of KClO 4 ​ in the final solid residue? (Atomic masses: K = 39 , Cl = 35.5 , O = 16 ). (Round off to nearest integer)

TestHub · Accepted Answer

Initial moles of KClO₃ = 12.25 g /122.5 g / m o l = 0.1 m o l . Mass of O₂ evolved = 12.25 g − 8.54 g = 3.71 g . Moles of O₂ = 3.71 g /32 g / m o l = 0.1159375 m o l . Let x be moles of KClO₃ decomposing via pathway (I) and y via pathway (II). x + y = 0.1 (Total moles of KClO₃) From (I): 2 K Cl O 3 ​ → 3 O 2 ​ , so moles of O₂ from (I) = ( 3/2 ) x . Thus, ( 3/2 ) x = 0.1159375 m o l ⇒ x = 0.07729167 m o l . Then y = 0.1 − x = 0.1 − 0.07729167 = 0.02270833 m o l . From (II): 4 K Cl O 3 ​ → 3 K Cl O 4 ​ + K Cl . Moles of KClO₄ formed = ( 3/4 ) y = ( 3/4 ) × 0.02270833 = 0.01703125 m o l . Molar mass of KClO₄ = 39 + 35.5 + 4 × 16 = 138.5 g / m o l . Mass of KClO₄ = 0.01703125 m o l × 138.5 g / m o l = 2.3608 g . Mass percentage of KClO₄ in residue = ( 8.54 g 2.3608 g ​ ) × 100% = 27.64% ≈ 27.6% .

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