posted by Deepa on .
Hi!! pls i need help!
1) Mg + Al2(So4)3
2)Fe + CuSO4
3)Mg + CuSO4
4)Zn + CuSO4
5)Mg + FeSO4
FOR ALL 5:
c) net ionic
I will do the first one in detail. The others follow the same pattern.
Look up the activity series in your text/notes/Google/etc.
If the METAL is above the metal ION, the metal will displace it; otherwise, there is no reaction.
3Mg + Al2(SO4)3 --> 3MgSO4 + 2Al
Mg is ABOVE Al in the activity series; therefore, the ion is displaced. These are single replacement type reactions. Then I balanced the equation.
Turn the balanced equation into an total ionic equation. I will write (aq), (s), (l) after each to show it is a solid, aqueous, pure material (that's l).
3Mg(s) + 2Al^3+(aq) + 3SO4^2-(aq) ==> 3Mg^+2(aq) + 3SO4^2-(aq) + 2Al(s)
The net ionic equation is obtained by canceling anything that appears on both sides of the equation. Here the SO4^2-(aq) appears on both sides so the net ionic equation is
3Mg(s) + 2Al^3+(aq) ==> 3Mg^2+(aq) + 2Al(s)