chemistry class

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)

A sample of 42.1 g MnO2 is added to a solution containing 50.5 g HCl.

What is the theoretical yield of Cl2?

If the yield of the reaction is 77.5%, what is the actual yield of chlorine?

Answer 1

Find the limiting reactant:

moles MnO₂ = 42.1 g MnO₂ x 1 mol/86.9 g = 0.484 moles MnO₂ (÷ 1 => 0.484)

moles HCl = 50.5 g HCl x 1 mol/36.5 g = 1.38 moles HCl (÷ 4 => 0.346)

HCL IS LIMITING

Find theoretical yield of Cl₂:

1.38 moles HCl x 1 mol Cl₂ / 4 mol HCl x 70.9 g Cl₂/mol Cl₂ = 24.5 g Cl₂

Find actual yield:

actual yield / theoretical yield (x100) = % yield

actual yield = theoretical yield x % yield

actual yield = 24.5 g x 0.775 = 19.0 g