Can someone help me balance these chemical equations and this radioactive decay question?

Question

___ I2 + ___ S2O32- 


___ Cu2+ + ___ NH3 


___ Mg2+ + ___ OH- 


___ Ba2+ + ___ SO42- 


How many years would it take for the mass of a 1.000g-sample of carbon-14 to be reduced to 0.125 g?

Answer 1

I2 + 2 S2O3^2- ---> S4O6^2- + 2 I-

Cu2+ + 4 NH3 ---> [Cu(NH3)4]^2+

Mg^2+ + 2 OH- ----> Mg(OH)2

Ba^2+ + SO4^2- ---> BaSO4

C-14's half-life is roughly 5700 years, and to go from 1.000 g to 0.125 g is three half lives (losing half of the remaining each cycle, so from 1.000 to 0.500 in the first half life, 0.500 to 0.250 in the second, etc.), so it will take 17,100 years.

Answer 2

Carbon 14 has a half life of ~5600 years

therefore.....
1 / 2 = 0.5 = 5600 years (1 half life)

1/2 / 2 = 1/4 = 0.25 = 11200 years (2 half lives)

1/4 / 2 = 1/8 = 0.125 = 16800 years (and 3 half lives)