Chemistry Forum

Hi,

I was asked to create the overall balanced equation for....

2Al(s) + 2koh(aq) + 6H₂O(l) = 2K[Al(OH)₄](aq) + 3H₂(g)
2K[Al(OH)₄](aq) + H₂SO₄(aq) = 2AL(OH)₃(s) + K₂SO₄(aq) + 2H₂)(l)
2Al(OH)₃(s) + 3H₂SO₄(aq) = 2Al^3+(aq) + 3SO₄^2-(aq) + 6H₂)(l)
K^+(aq) + Al^3+(aq) + 2SO₄^2-(aq) + 12H₂O(l) = KAl(SO₄)₂ x 12H₂O(s)

And I got,

2Al(s) + 2KOH(aq) + 4H₂SO₄(aq) + K(aq) + 10H₂O(l) = 3H₂(g) + K₂SO₄(aq) + Al^3+(aq) + SO₄^2-(aq) + KAl(SO₄)₂ x 12H₂O(s)

I was then asked to find the %yield so I figured I needed to find the limiting reagent so stoichiometry, but I am confused on how to balance this gigantic equation in order to find the limiting reagent.

Please provide a solution for this as I am totally lost in this problem

Thanks and sorry for the gigantic equation.

The final equation is

2 Al + 2 KOH + 4 H₂SO₄ + 22 H₂O => 2( KAl(SO₄)₂ * 12 H₂O) + 3 H₂

Now it depends how much you have from each edukt, then you can calculate the yield and the limiting componente.

Thank you so much.

You wellcome