A 500-g sample of Al2 (SO4) 3 is reacted with 450 g of Ca (OH) 2. A total of 596 g of CaSO4 is produced. What is the limiting reagent in this reaction, and how many moles of excess reagent are unreacted?

The chemical reaction is as follows:

Al2 (SO4) 3 + 3Ca (OH) 2 → 3CaSO4 + 2Al (OH) 3

Explanation:

1mol of Al2 (SO4) 3 will react with 3mol Ca (OH) 2 to produce 3mol CaSO4 and 2 mol Al (OH) 3.

First we have to find the number of moles of Al2 (SO4) 3:

Number of moles = Mass / Molar mass

Mass of Al2 (SO4) 3 = 500g

Molar mass of Al2 (SO4) 3 = 342.15 g/mol

Number of moles = 500 / 342.15

Number of moles = 1.461 mol Al2 (SO4) 3

Multiplying the coeffecient of Ca (OH) 2 with 1.461:

= 3*1.461 = 4.383 mol Ca (OH) 2

Now we have to find the number of moles of Ca (OH) 2:

Mass of Ca (OH) 2 = 450g

Molar mass of Ca (OH) 2 = 74.09 g/mol

Number of moles = 450 / 74.09

Number of moles = 6.074 mol Ca (OH) 2

We need 4.383mol to react completely with the Al2 (SO4) 3, so the Ca (OH) 2 is in excess, and the Al2 (SO4) 3 is the limiting reactant.

Excess unreacted: 6.074-4.383 = 1.69mol Ca (OH) 2 unreacted.