Question

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide:You add 40.7 g of MnO2 to a solution containing 42.7 g of HCl.(a) What is the limiting reactant?(b) What is the theoretical yield of Cl2?(c) If the yield of the reaction is 81.5%, what is the actual yield of chlorine?

Answer 1

:  4 HCl + MnO2 → MnCl2 + 2 H2O + Cl2 

(33.7 g MnO2) / (86.93691 g MnO2/mol) = 0.38764 mol MnO2 
(45.3 g HCl) / (36.4611 g HCl/mol) = 1.2424 mol HCl 

(a) 
1.2424 moles of HCl would react completely with 1.2424 x (1/4) = 0.3106 mole of MnO2, but there is more MnO2 present than that, so MnO2 is in excess and HCl is the limiting reactant. 

(b) 
(1.2424 mol HCl) x (1 mol Cl2 / 4 mol HCl) x (70.9064 g Cl2/mol) = 22.0 g Cl2 

(c) 
(77.7% of 22.0 g Cl2) = 17.1 g Cl2