Close but no cigar.
Full equation is
MgCl2(aq) + 2NaOH(aq) ==> Mg(OH)2(s) + 2NaCl(aq)
Those (aq) reagents are put in ionic form.
Mg^2+(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2OH^-(aq) ==> Mg(OH)2(s) + 2Na^+(aq) + 2Cl^-(aq)
Then cancel 2Cl^- and 2Na^+ since they appear on both sides and you are left with
Mg^2+(aq) + 2OH^-(aq) ==> Mg(OH)2(s)
What is the net ionic equation of reaction of MgCl2 with NaOH?
I got the equation and then balanced it and then omitted the spectator ions and ended up with:
Mg(OH)2(aq)--->Mg(OH) (s)
I wanted to confirm to see if that was correct, thanks.
1 answer
1 answer