A student in a lab adds 20.0 mL of 0.150M Al2(SO4)3 solution to 30.0 mL of 0.200M BaCl2 solution, a precipitate forms. Which is the limiting reactant?

Question

Limiting reactant question

Answer 1

The results of the reaction are AlCl₃ and BaSO₄.  Of the two, only BaSO₄ is insoluble, so that is the precipitate.

 

First let's determine the balanced equation:

 

Al₂(SO₄)₃ + BaCl₂ ---> AlCl₃ + BaSO₄

Al₂(SO₄)₃ + 3BaCl₂ ---> 2AlCl₃ + 3BaSO₄

 

So 1 mole of Al₂(SO₄)₃ will react with 3 moles of BaCl₂.

 

Now we have to determine the number of moles there are of each reactant.

 

20 mL of 0.150M Al₂(SO₄)₃ is .02 * .15 = .0030 moles

 

30 ml of 0.200M BaCl₂ is .03 * .20 = 0.0060 moles

 

To react with .003 moles of Al₂(SO₄)₃, there would have to be .009 moles of BaCl₂, but there are only 0.006 moles.  Therefore, there isn't enough BaCl₂ and that is the limiting reactant.