Question #cecb0

Let us denote each components using the following letters
#A = (NH_4)_2SO_4#
#B = NH_4^+#
#C = SO_4^(2-)#
#D = NH_3(aq)#
#E = H^+#
#F = H_2O#
#G = HSO_4^-#
#H = OH^-#

Then we have
#A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}}2B+C#
#B \underset{k_4}{\stackrel{k_3}{\rightleftharpoons}} D+E#
#C+F \underset{k_6}{\stackrel{k_5}{\rightleftharpoons}}G+H#

If you are writing a component wise mass balance then start with components that take part in a single reaction

#(dC_A)/dt = k_2C_B^2C_C-k_1C_A#
#(dC_D)/dt = k_3C_B-k_4C_DC_E#
#(dC_E)/dt = k_3C_B-k_4C_DC_E#

Generally water molecule is present in excess. Hence we don't write a balance for that. And hence the third reaction is generally treated as pseudo first order reaction.

#(dC_G)/dt = k_5C_C-k_6C_GC_H#
#(dC_H)/dt = k_5C_C-k_6C_GC_H#

Now onto the other 2 components

#(dC_B)/dt = k_1C_A-k_2C_B^2C_C-k_3C_B+k_4C_DC_E#
#(dC_C)/dt = k_1C_A-k_2C_B^2C_C - k_5C_c+k_6C_GC_H#

Then for each component do a mass balance as
#v_0(C_i)_o - v(C_i) +V(r_i) = V(dC_i)/dt#

where #i \in {A,B,C,D,E,G,H}#

If you are writing an overall balance then we have the overall reaction as
#A+F \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} B+D+E+G+H#

Then we can write balances for A,B,D,E,G,H given we have #k_1# and #k_2#.