posted by jake on .
I need help finding the net and total ionic for this reation. i know the steps but i keep getting the wrong answer.
Al2(SO4)3 (aq) + 6 NaOH (aq) ----> 2 Al(OH)3 (s) +3 Na2SO4 (aq)
Al2(SO4)3 (aq) + 6 NaOH (aq) ---->
2Al(OH)3 (s) +3 Na2SO4 (aq)
1. Separate into ions. Solids stay as solids, aqueous solution are separated. (Gases, if you had one, would be written as the molecule, also).
2Al+3(aq) + 3SO4^-2(aq) + 6 Na^+(aq) + 6 OH^-(aq) ==> 2Al(OH)3(s) + 6Na^+(aq) + 3SO4^-2(aq)
Now cancel the ions common to both sides. 3SO4^-2 cancel. 6Na^+ cancel. You are left with
2Al^+3(aq) + 6OH^-(aq) ==> 2Al(OH)3(s)
Check my work. This is meticulous.
thanks.. i had probmels with the subscripts and coefficents. i didn't know you had to muliply it