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 %

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.