Finding the theoretical yield and actual yield?

Question

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide.

4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

A sample of 34.9 g MnO2 is added to a solution containing 43.3 g HCl.

What is the theoretical yield of Cl2?

theoretical yield:

If the yield of the reaction is 80.3%, what is the actual yield of chlorine?

actual yield:

Answer 1

4HCl(aq) + MnO₂(s) ⟶ MnCl₂(aq) + 2H₂O(l) + Cl₂(g) ... balanced equation

Whenever you are given the amounts of BOTH reactants, you must first find 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.

For HCl we have 43.3 g HCl x 1 mol HCl / 36.5 g = 1.19 mols HCl (÷4->0.29 )

For MnO₂ we have 34.9 g MnO₂ x 1 mol MnO₂ / 86.9 g = 0.402 mol (÷1->0.402)

Since 0.29 is less than 0.40, the HCl is the limiting reactant, and the mols of HCl (1.19) will determine the theoretical yield of Cl2.

Theoretical yield of Cl₂ = 1.19 mol HCl x 1 mol Cl₂ / 4 mol HCl x 71 g / mol = 21.1 g Cl₂

Actual yield: (80.3% )(21.1 g) = 16.9 g Cl₂