[prncess23]

Okay...im pretty good at balancing equation but i dont know why i cant do this one.

seems like something is missing in the reaction given.

The question states....
Balance the following equation:

[gfunk]

Firstly, what kind of chemical process is this?

[prncess23]

redox

[ARGOS++]

Dear prncess23;

How about to do it strong acidic like the "Bechamp-Reduction"?

Good Luck!
                    ARGOS⁺⁺

[AWK]

Use HCl on the left side. Sn(IV) on the right side is doubtful, and the final product will be a salt of this amine.

[prncess23]

so.
it will be 1-methyl-3,5-dinitrobenzene + HCl +Sn --->2,4-diaminotolulene + Sn⁺⁴

[ARGOS++]

Dear prncess23;

Yes,  -  but now you have to balance it!
And finally you can make the salt of the di-amine.

Good Luck!
                    ARGOS⁺⁺

[prncess23]

ok..so i tried balancing the equation and this is what i got....i dont think its right...cuz the atoms arent balanced...

what am i doing wrong? i am good at balancing redox equations but it looks like something is not given in the equation given in the problem.

half reactions:
C7H6N2O4 --> C7H10N2
Sn --> Sn+4
------------------------------------------------
2(C7H6N2O4 + 6H2O --> C7H10N2 + 4H2O + 6H3O + 6e-)
3(Sn + 4e- --> Sn+4)

2C7H6N2O4 + 12H2O --> 2C7H10N2 + 8H2O + 12H3O + 12e-
3Sn + 12e- --> 3Sn+4

Overall equation:
2C7H6N2O4 + 12H2O + 3Sn --> 2C7H10N2 + 8H2O + 12H3O + 3Sn+4

[UG]

3(Sn + 4e- --> Sn+4)

2C7H6N2O4 + 12H2O --> 2C7H10N2 + 8H2O + 12H3O + 12e-
3Sn + 12e- --> 3Sn+4

Those are wrong. Electrons on the wrong side.

[prncess23]

half reactions:
C7H6N2O4 --> C7H10N2
Sn --> Sn+4
------------------------------------------------
2(C7H6N2O4 + 6H2O --> C7H10N2 + 4H2O + 6H3O + 6e-)
3(Sn --> Sn+4 + 4e-)

2C7H6N2O4 + 12H2O --> 2C7H10N2 + 8H2O + 12H3O + 12e-
3Sn --> 3Sn+4 + 12e-

Overall equation:
2C7H6N2O4 + 12H2O + 3Sn --> 2C7H10N2 + 8H2O + 12H3O + 3Sn+4 + 12e-

that still isnt right though.

[UG]

That is because this half-equation is incorrect:

2C₇H₆N₂O₄ + 12H₂O  2C₇H₁₀N₂ + 8H₂O + 12H₃O⁺ + 12e^-

There are only 36 hydrogens on the LHS but 60 on the RHS, also the water molecules on either side should cancel each other out. The electrons are also on the wrong side.

[prncess23]

so is this right?

2C7H6N2O4 + 12e-  2C7H10N2 +  + 12H3O+

[UG]

2C₇H₆N₂O₄ + 12e^- 2C₇H₁₀N₂ +  + 12H₃O⁺

Now there are 12 hydrogens on the LHS and 56 on the RHS, also the charges on either side are not balanced, -12 on the left and +12 on the right. Why do you use H₃O⁺? Using H⁺ would make things a lot simpler, I believe 

[prncess23]

2C₇H₆N₂O₄ + 12e-  2C₇H₁₀N₂ +  2H⁺

what about the Sn?

[UG]

2C₇H₆N₂O₄ + 12e-  2C₇H₁₀N₂ +  2H⁺

Come on mate, that's still not balanced!   
You've got 12 hydrogen on the LHS and 22 on the other. 
Neither is the charge balanced. Maybe you need a quick jog of your memory on how to do this.
I posted this earlier in the day: http://www.chemicalforums.com/index.php?topic=33050.msg126713#msg126713

Good Luck!