A 500-g sample of Al2(SO4)3 is reacted with 450 g of Ca(OH)2. A total of 596 g of calcium sulfate is produced in the laboratory. What is the percent yield?

The percent yield is 99.9 %

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 %

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8