Question

chlorine gas can be prepared in the lab by the reaction of hydrochloric acid with manganese (IV) oxide.
you add 34.9g MnO2 to a solution containing 45.7 g HCl

A) what is the limiting reactant?

B) what is the theoretical yield of Cl2 in grams?

C) if the yield of the reaction is 78.3%, what is the actual yield if chlorine ?

Answer 1

The balance reaction is as follows:

4 HCl + MnO2 → MnCl2 + 2 H2O + Cl2

Given that

34.9 g MnO2

45.7 g HCl

First calculate the number of mole of both reactants as follows:

(34.9 g MnO2) / (86.93691 g MnO2/mol) = 0.40 mol MnO2
(45.7 g HCl) / (36.4611 g HCl/mol) = 1.25 mol HCl

(a)
1.25 moles of HCl would react completely with 1.25 x (1/4) = 0.3125 mole of MnO2, but there is more MnO2 present than that, so MnO2 is in excess and HCl is the limiting reactant.

HCl is limiting agent due to following reasons:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

(b)
(1.25 mol HCl) x (1 mol Cl2 / 4 mol HCl) x (70.9064 g Cl2/mol) = 22.16 g Cl2

(c)
(78.3% of 22.16 g Cl2) = 17.35 g Cl2 actually