Question

What quantity of ammonium sulphate is necessary for the production of nh3 gas which will neutralise a solution containing 292g hcl?

Answer 1

(NH4)2SO4 + 2NaCl → Na2SO4 + NH3 + 2HCl

according to reaction

one mole of(NH4)2SO4  = one mole of NH3 = two mole of HCl

292 g Hcl = 292 /36.5 mole of Hcl = 8 moles of Hcl

2 moles of HCl produced from one mole of (NH4)2SO4

8 moles of HCl produced from 4 moles of (NH4)2SO4

amount of (NH4)2SO4 = 4 *molecular weight of (NH4)2SO4

                                      = 4 *132.14 = 528 g

I hope its helpful .

thanks!!

Answer 2

Answer:

528 g

Explanation:

Eq. of (NH4)2SO4= Eq. of HCl.

Let, the amount of (NH4)2SO4 needed is x g. 

Then, 132/2x=36.5292.

Hence, x=528 g.