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)

You add 38.3 g of MnO2 to a solution containing 47.3 g of HCl.

a)What is the limiting reaction?

b) What is the theoretical yield of Cl2?

c) If the yield of the reaction is 80.7%, what is the actual yield of chlorine?

Answer 1

Ans

a) 38.3 g of MnO2 = 38.3 / 86.9 = 0.44 moles

47.3 HCl = 47.3 / 36.46 = 1.3 moles

1 mole MnO2 requires 4 moles of HCl

0.44 moles of MnO2 will require 0.44 x 4 = 1.76 moles of HCl

So HCl is the limiting reagent.

b) 1 mole MnO2 makes 1 mole of Cl2

So 0.44 moles of MnO2 will make 0.44 moles of Cl2

Theoretical yield of Cl2 = 0.44 x 70.906

= 31.2 g

c) Percent yield = (actual yield / theoretical yield) x 100

actual yield = (80.7 / 100) x 31.2

Actual yield = 25.2 g