33 г                                                               Х л

       (NH4)2SO4      +    2NaOH      =    Na2SO4 + 2NH3 +   2H2O

 n =  1 моль                                                           n = 2 моль

Mr = 132 г/моль                                                    Vm = 22,4 л/моль

m = 132 г                                                                V = 44,8 л

33 г (NH4)2SO4     -     Х л NH3

132 г (NH4)2SO4   -    44,8 л NH3

V(NH3) = 33 * 44,8 / 132 = 11,2 л