Question

Find the volume of ammonia required for the preparation of 66gm of ammonium sulphate according to the reaction? 2NH3+H2SO4=(NH4)2SO4?

Answer 1

Well....we need approx. a `67*mL` volume of `NH_3(aq)`..

Explanation:

`underbrace(2NH_3(aq) + H_2SO_4(aq))_(98.08*g*mol^-1+34.06*g*mol^-1) rarr underbrace((NH_4)_2SO_4(aq))_(132.14*g*mol^-1)`

Now clearly, garbage in equals garbage out; that is mass is conserved. And here we finish with a `66*g` mass of products..i.e. `(66.0*g)/(132.14*g*mol^-1)=0.500*mol` WITH RESPECT to ammonium sulfate....and thus there were `1*mol` with respect to ammonia. Now ammonia is a room temperature gas, but here, likely, we started with an aqueous solution of ammonia...and conc. ammonia (in water) is typically `15*mol*L^-1`...

And so we need a volume of `(1*mol)/(15*mol*L^-1)xx1000*mL*L^-1=66.5*mL` with respect to aqueous ammonia..