Let us denote each components using the following letters
A = (NH_4)_2SO_4A=(NH4)2SO4
B = NH_4^+B=NH+4
C = SO_4^(2-)C=SO2−4
D = NH_3(aq)D=NH3(aq)
E = H^+E=H+
F = H_2OF=H2O
G = HSO_4^-G=HSO−4
H = OH^-H=OH−
Then we have
A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}}2B+CAk1⇌k22B+C
B \underset{k_4}{\stackrel{k_3}{\rightleftharpoons}} D+EBk3⇌k4D+E
C+F \underset{k_6}{\stackrel{k_5}{\rightleftharpoons}}G+HC+Fk5⇌k6G+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_AdCAdt=k2C2BCC−k1CA
(dC_D)/dt = k_3C_B-k_4C_DC_EdCDdt=k3CB−k4CDCE
(dC_E)/dt = k_3C_B-k_4C_DC_EdCEdt=k3CB−k4CDCE
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_HdCGdt=k5CC−k6CGCH
(dC_H)/dt = k_5C_C-k_6C_GC_HdCHdt=k5CC−k6CGCH
Now onto the other 2 components
(dC_B)/dt = k_1C_A-k_2C_B^2C_C-k_3C_B+k_4C_DC_EdCBdt=k1CA−k2C2BCC−k3CB+k4CDCE
(dC_C)/dt = k_1C_A-k_2C_B^2C_C - k_5C_c+k_6C_GC_HdCCdt=k1CA−k2C2BCC−k5Cc+k6CGCH
Then for each component do a mass balance as
v_0(C_i)_o - v(C_i) +V(r_i) = V(dC_i)/dtv0(Ci)o−v(Ci)+V(ri)=VdCidt
where i \in {A,B,C,D,E,G,H}i∈{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+HA+Fk1⇌k2B+D+E+G+H
Then we can write balances for A,B,D,E,G,H given we have k_1k1 and k_2k2.
