Question

1)how many moles of Al are in 1mole of KAI(SO4)2 ---->12H20?

2)what is the molar mass of potassium aluminum sulfate dodecahydrate?

3)if you have 1.20g of Al, how many grams of potassium aluminum sulfate dodecahydrate should you expect to produce if you have no other limiting reactants?

4)using your data from question #4 if you actually produce 6.6g of potassium aluminum sulfate dodecahydrate what is your percentage yield?

5)based on your results from question #4 if your percentage yield was 112% what mass of potassium aluminum sulfate dodecahydrate would you recover?

Answer 1

(1) From the formula KAI(SO₄)₂ .12H₂O , 1 mole of KAI(SO₄)₂ .12H₂O contains 1 mole of Al

(2) Molar mass of KAI(SO₄)₂ .12H₂O ( condensed formula is KAlS₂H₂₄O₂₀)= sum of the atomic numbers of all atoms

= (At.mass of K)+(At.mass of Al)+(2xAt.mass of S)+(24xAt.mass of H)+(20xAt.mass of O)

= 39+27+(2x32)+(24x1)+(20x16)

= 474 g/mol

(3) 1 mole= 474 g of potassium aluminum sulfate dodecahydrate contains 1mol=27 g of Al

M g of potassium aluminum sulfate dodecahydrate contains 1.20 g of Al

M = (1.20x474) / 27

    = 21.1 g of potassium aluminum sulfate dodecahydrate ---> This is the theoretical yield

(4) pecentage yield = ( actual yield / theoretical yield) x100

                             = ( 6.6 / 21.1) x100

                             = 31.3 %

(5) Given percentage yield is 112%

pecentage yield = ( actual yield / theoretical yield) x100

                112 = ( actual yield / 21.1) x100

So actual yiled = 23.6 g