Answer:
The percent yield is 99.9 %
Step-by-step explanation:
Step 1: Data given
Mass of a sample Al2(SO4)3 = 500 grams
Mass of Ca(OH)2 = 450 grams
Mass of CaSO4 produced = 596 grams
Molar mass Al2(SO4)3 = 342.15 g/mol
Molar mass Ca(OH)2 = 74.09 g/mol
Step 2: The balanced equation
Al2(SO4)3 + 3Ca(OH)2 → 2Al(OH)3 + 3CaSO4
Step 3: Calculate moles
Moles = mass / molar mass
Moles Al2(SO4)3 = 500 grams / 342.15 g/mol
Moles Al2(SO4)3 = 1.46 moles
Moles Ca(OH)2 = 450 grams / 74.09 g/mol
Moles Ca(OH)2 = 6.07 moles
Step 4: Calculate the limiting reactant
For 1 mol Al2(SO4)3 we need 3 moles Ca(OH)2 to produce 2 moles Al(OH)3 and 3 moles CaSO4
Al2(SO4)3 is the limiting reactant. It will completely be consumed (1.46 moles). Ca(OH)2 is in excess. There will react 3*1.46 = 4.38 moles
There will remain 6.07 - 4.38 = 1.69 moles
Step 5: Calculate moles CaSO4
For 1 mol Al2(SO4)3 we need 3 moles Ca(OH)2 to produce 2 moles Al(OH)3 and 3 moles CaSO4
For 1.46 moles Al2(SO4)3 we'll have 3*1.46 moles = 4.38 moles CaSO4
Step 6: Calculate mass CaSO4
Mass CaSO4 = 4.38 moles * 136.14 g/mol
Mass CaSO4 = 596.3 grams
Step 7: Calculate percent yield
Percent yield = (actual yield / theoretical yield) * 100%
Percent yield = (596 grams / 596.3 grams ) *100 %
Percent yield = 99.9 %
The percent yield is 99.9 %