Asked • 09/26/18

Percent Yield of Alum Lab

I've recently did a lab to turn aluminum foil into alum using KOH and Sulfuric Acid.

I need to find the percent yield of alum, but don't how to figure out which information I need.

Here is what I have:

1.01 g of Al (starting material)
12.77 g of alum (final product)
50 mL of KOH (used in reaction)
21 mL of Sulfuric Acid (used in reaction)

All the reactions that took place in chemical formulas (unbalanced):
Al(s) + KOH(aq) + H2O(l) = KAl(OH)4(aq) + H2(g)
KAl(OH)4(aq) + H2SO4(aq) = K2SO4(aq) + Al(OH)3(s) + H2O(l)
Al(OH)3(s) + H2SO4(aq) = Al2(SO4)3(aq) + H2O(l)
Al2(SO4)3(aq) + K2SO4(aq) + H2O(l) = KAl(SO4)2 * 12H2O(s)

1 Expert Answer


Daisy S.

Would I have to convert all the coefficients of H2O with each mol of my other substances?


J.R. S.

Sorry. I don’t understand your question. If I think I know what you’re asking, then my answer is only the H2O that appears as a reactant in the first and last  equation. But I’m not sure I understand the question. 


Daisy S.

So once I balance out all my equations, and found the moles of each compound, would I need to divide coefficients of Reaction 1 and 4?


Daisy S.

Nevermind, I figured it out! Thank you for the help!


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