Question

What quantity of ammonium sulphate is necessary for the production of NH3 gas sufficient to
neutralize a solution containing 292 gm of HCI?
(A) 272 gm
(B) 408 gm
(C) 528 gm
(D) 1056 gm​

Answer 1

Answer:

Production of ammonia from Ammonium sulphate (NH4)2SO4 + 2NaOH →2NH3 +2H2O +Na2SO4 2) Neutralisation : HCl + NH3 →NH4Cl 1 mole of HCl neutralises = 1 mole of NH3 No. of moles of HCl = 292 g /36.5 g/mol = 8 moles No. of moles of NH3 neutralised by 8 moles of HCl = 8 moles Now from equation 1 we know 2 moles of ammonia is produced by = 1 mole of ammonium sulphate 1 mole of ammonia is produced by = 1/2 mole of ammonium sulphate 8 moles of ammonia is produced by = (1/2) x 8 = 4 moles of Ammonium sulphate Molar mass of ammonium sulphate = 132.14 g/mol mass of 4 mole of ammonium sulphate is = 4 x 132.14 = 528.56 g

so the answer will be ..(c) 528 gm