how to solve this...

[zain tariq]

(c) For each of the following equations balance the equation by adding appropriate numbers before each species.

(i) ........ MnO^–4 + ........ SO2 + ........ H2O → ........ Mn^+2 + ........ So4^- 2 + ........ H+



(ii) ........ Cr2O7^-2 + ........ NO2 + ........ H+ → ........ Cr^+3 + ........ NO^–3 + ........ H2O



is there any quick way of balancing this cause the problem is to balance charges alongwith numbers which is time consuming... :unknown:

 

[destined007]

This is a short way:
look at the two reduction equations for the reaction from the data booklet:
MnO4– + 8H+ +5e–⇌ Mn2+ + 4H2O
SO42– + 4H+ + 2e–⇌ SO2 + 2H2O

balance the number of electrons for both half equations:
2x(MnO4– + 8H+ +5e–⇌ Mn2+ + 4H2O)
5x(SO42– + 4H+ + 2e–⇌ SO2 + 2H2O)

the equations become
2MnO4– + 16H+ +10e–⇌ 2Mn2+ + 8H2O
5SO42– + 20H+ + 10e–⇌ 5SO2 + 10H2O
electron are cancel out
Now subtract the coefficient of particles that are same in both half equations, ie, H+ and H20.
H+= 20-16=4
H20= 10-8=2

DONE! put the number in the respective field.
2MnO^–4 + 5SO2 + 2H2O → 2Mn^+2 + 5SO4^- 2 + 4H+

Now do for the 2nd equation