Question

Al2(SO4)3+6KOH=2Al(OH)3+3K2SO4 if he added 55.5 g of Aluminum Sulfate to 55.5 g of Potassium hydroxide and produced 18.65 g of Aluminum Hydroxide what is the percent yield?

Answer 1

Al2(SO4)3 + 6 KOH ------------>  2Al(OH)3 + 3K2SO4

first calculate the limiting reagent.

342.15 g Al2(SO4)3 reacts with 6 x 56.1 g KOH

55.5 g Al2(SO4)3 reacts with 55.5 x 6 x 56.1 / 342.15 = 54.6 g KOH

but we have 55.5 g KOH so KOH is exess reagent.

Al2(SO4)3 is limiting reagent.

342.15 g Al2(SO4)3 forms 2 x 78 g Al(OH)3

55.5 g Al2(SO4)3 forms 55.5 x 2 x 78 / 342.15 = 25.5 g

therotical yield = 25.5 g

actual yield = 18.65 g

% yield = (actual yield / therotical yield) x 100

% yield = (18.65 / 25.5) x 100

% yield = 73.14 %