Percent Yield of Alum Lab

Question

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) + H_2(g)
KAl(OH)_4(aq) + H₂SO_4(aq) = K₂SO_4(aq) + Al(OH)_3(s) + H₂O_(l)
Al(OH)_3(s) + H₂SO_4(aq) = Al₂(SO₄)_3(aq) + H₂O_(l)
Al₂(SO₄)_3(aq) + K₂SO_4(aq) + H₂O_(l) = KAl(SO₄)₂ * 12H₂O_(s)

Answer 1

You need to find out which is the limiting reactant.  Is it Al(s), KOH(aq) or H2SO4(aq).  To do this, you need to balance all the equations.  Can you do that?

 

After balancing, find moles of each.  For Al, it will be 1.01 g/atomic mass of Al

For KOH, it will be 0.050 Lof KOH x the molar concentration (you didn't provide this information)

For H2SO4 it will be 21 ml x density / molar mass of H2SO4 (you didn't provide the density). Or if you are given the molarity, you can find moles using that value (0.021 L x mole/L = moles)

 

For H2O it will be moles determined in the balanced equation.

Once you have moles of each, divide those values by the corresponding coefficients in the balanced equation that uses that substance. The one that has the lowest value is limiting.

 

Based on the limiting reactant, you can then find the theoretical moles of KAl(SO4)2*12H2O

From moles, you can find grams by using the molar mass of the alum.

Finally, for % yield, it will be actual yield (12.77g) divided by the theoretical yield (x100%).