Question

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese...

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?

Homework Answers

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

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