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 43.1 g of MnO₂ to a solution containing 41.7 g of HCl.

What is the limiting reactant?

What is the theoretical yield of Cl₂?

If the yield of the reaction is 71.1% what is the actual yield of chlorine?

Answer 1

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

mol of MnO2 = mass/MW = 43.1/86.9368 = 0.49576243 mol MnO2

mol of HCl = mass/MW = 41.7/36 = 1.15833 mol of HCl

ratio is 4:1

1.15833 mol of HCl need = 1.15833 /4 = 0.2895825 mol of MnO2 which we DO have, therefore HCl is limiting

theretical yield of Cl2

4 mol of HCl: 1 mol of CL2

then

1.15833 /4 = 0.2895825 mol of Cl2 may be produced

yield in mass = mol*MW = 0.2895825*70 = 20.270 g of Cl2

if only 71.1%

then

yield = 20.270 *71.1/100 = 14.41 g of Cl2