Question

10 g sample of 'gas liquor' (NH4^+ salt) is boiled with NaOH and the resulting NH3 is passed into 60 ml of 0.9 H2SO4. Excess H2SO4 required 10cm^3 of 0.40 N NaOH. What is the ℅ of NH3 in gas liquor?

Answer 1

me of H2SO4 reacted with gas liquor =( 60 x 0.9 ) - ( 10 x 0.40) = 1.4 me = me of NH3 in the gas liquor =  w/17 /3
hence w= 1.4 x 17/3 = 7.93 g 
hence % of ammonia in gas liquor = 7.93/10  x 100 = 79.3% 

Answer 2

Answer:

Explanation:

Sample of gas liquor = 10g

Resulting quantity of the solution = H2SO4 = 60ml

Excess H2SO4 = 10cm³ of 0.40 N NaOH

Wt of H2SO4= 50 meq. =  eq. wt. of NH3

Therefore, mass of ammonia = 50 meq x 50/1000

= 0.85 gm.

Thus the reaction equations will be -

2NH3 + H2SO4 --> (NH4)2SO4  

from mole to mole analysis,  on solving we will again get -

50/1000 x 17

= 0.85 g.  

This, is so as one mole of H2SO4 reacts with two moles of ammonia.