What mass of aluminum hydroxide is produced?

Question

A student in a chemistry lab mixes 13.8g of aluminum sulfate with 2.04g calcium hydroxide to produce aluminum hydroxide and calcium sulfate.

The unbalanced reaction is: Al₂(SO₄)₃ + Ca(OH)₂  Al(OH)₃+ CaSO₄

Answer 1

First, make sure you have a correctly balanced equation:

Al₂(SO₄)₃ + 3Ca(OH)₂  2Al(OH)₃+ 3CaSO₄ .. balanced equation

Next, since we are given the amounts of BOTH reactants, we must determine which reactant is limiting. One way to do this is to simply divide the MOLES of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For Al₂(SO₄)₃: 13.8 g x 1 mol / 342.2 g = 0.0403 mols (÷1->0.04)

For 3Ca(OH)₂: 2.04 g x 1 mol / 74.09 g = 0.0275 mols (÷3->0.009)

Since 0.009 is less than 0.04, 3Ca(OH)₂ is limiting and the moles of 3Ca(OH)₂ (0.0275 mols) will determine how much product, aluminum hydroxide, can be produced.

Theoretical yield of Al(OH)₃ = 0.0275 mols Ca(OH)₂ x 2 mol Al(OH)₃ / 3 mol Ca(OH)₂ x 78.0 g / mol = 1.43 g Al(OH)₃