Diiodozinc I2Zn
products...>>>> BaI2 + Zn(OH)2
Notice Zn++ does not change valence, nor does I-
zinc hydroxide will form a precipate.
Net ionic
2(OH)- + Zn++ >>> Zn(OH)2
Confused on how to find the reactants in these two compounds
Ba(OH)2 (aq) + I2Zn (aq) -> ??
Additionally, I am not really sure how to find the ionic equation for this.
1 answer