What mass of ammonia is required to give a #264*g# mass of #"ammonium sulfate"#?

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Answer 1 · 2017-08-02T16:53:49

Approx. #68*g# of ammonia are required............

We need (i) a stoichiometric equation.......

#2NH_3(aq) + H_2SO_4(aq) rarr (NH_4)_2SO_4(aq)#

and (ii) equivalent quantities of #"ammonium sulfate"#

#"Moles of ammonium sulfate"=(264*g)/(132.14*g*mol^-1)=2*mol#.

And thus we need a #4*mol# quantity of ammonia (why #4 *mol#?).

And so...............

#"mass of ammonia"=4*molxx17.03*g*mol^-1=??*g#