Posted by Tiffany on Sunday, March 25, 2012 at 3:28pm.
75% of the 4 g mixture will be Al(OH)3 or 3 g.
Al(OH)3 + OH^- ==> Al(OH)4^-
Al(OH)4^- + 2H2SO4 ==> Al^3+ + 4H2O + 2SO4^2-
What about the iron (II) oxide? and the KOH?
If the iron oxide is in solution then it will form Fe(OH)3 when KOH is added but adding excess KOH will NOT dissolve Fe(H)2 as it does Al(OH)3. That's how iron and aluminum are separated industrially; i.e., ppt both as the hydroxide, add excess to dissolve the Al, separate, then add acid to reppt the Al. But the way your problem is worded, the iron oxide is never placed in soln and I don't think appreciable iron oxide will dissolve in KOH. I think all you are doing is dissolving the Al(OH)3 in KOH and filtering the solution to remove the iron oxide.
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