J.R. S. answered 03/11/22
Ph.D. University Professor with 10+ years Tutoring Experience
2A + 3B ==> 2C ... balanced equation
This problem requires you to find the limiting reactant in each case. Any time the amounts of BOTH reactants are given to you, you will need to do this. One easy way to find the limiting reactant is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation and whichever value is less is the limiting reactant (see examples below)
3 mol A and 6 mol B
Find limiting reactant:
For A we have 3 mols (÷ 2 ->1.5)
For B we have 6 mols (÷3->2)
So A is the limiting reactant.
Next, to find the theoretical yield of C, we use the MOLES of the limiting reactant and dimensional analysis:
3 mols A x 2 mols C / 3 mols A = 2 moles C is the theoretical yield
4 mol a and 4 mol B
Find limiting reactant:
For A we have 4 mol (÷ 2->2)
For B we have 4 mol (÷3->1.3)
So B is the limiting reactant
Next, to find theoretical yield of C, we use the MOLES of the limiting reactant and dimensional analysis:
4 mols B x 2 mols C / 3 mols B = 2.67 mols C is the theoretical yield