Question

A solution of pure ferric sulphate containing 0.140 g of ferric ions is treated with excess of barium hydroxide solution. Total weight of the precipitate will be.

(a) 0.87 g 

(b) 1.14 g 

(c) 0.25 g 

(d) 0.56 g

Answer 1

Correct option: (b) 1.14 g

Explanation: 

∴ moles of pure ferric sulphate taken = 0.00125 moles

As per the reaction Fe(OH)₃ and BaSO₄ both will precipitate out.

So, total amount of precipitate = amount of Fe(OH)₃ + amount of BaSO₄

Amount of Fe(OH₃) = 2 × 0.00125 × 107 = 0.2675 gm

Amount of BaSO₄ = 3 × 0.00125 × 233 = 0.87375 gm

Total weight of precipitate = Fe(OH)₃ + BaSO₄ = 0.2675 + 0.8737 = 1.14 gm.