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

Question

7. 4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

A sample of 40.3 g MnO2 is added to a solution containing 50.5 g HCl.

(a) If the yield of the reaction is 87.1% what is the actual yield of chlorine?

(b) Identify the theoretical yield of Cl2

Answer 1

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

First thing...find the limiting reactant:

40.3 g MnO2 x 1 mol MnO2 / 86.9 g x 1 mol Cl2 / mol MnO2 x 70.9 g Cl2/mol = 32.9 g Cl₂

50.5 g HCl x 1 mol HCl / 36.5 g x 1 mol Cl2 / 4 mol HCl x 70.9 g Cl2/mol = 24.5 g Cl₂ = Theoretical yield

percent yield = actual yield / theoretical yield

actual yield = percent yield x theoretical yield = 0.871 x 24.5 g = 21.4 g Cl₂ actual yield