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 34.3 g of MnO2 to a solution containing 48.5g of HCl. what is the limiting reactant? What is the theoretical yield? If yield is 77.9%, what us the actual yield of chlorine?

Answer 1

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

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Molar mass of MnO₂ is = 55+2(16) = 87 g/mol

Molar mass of Cl₂ = 2x 35.5 = 71 g/mol

From the balanced equation ,

4 moles of HCl reacts with 1 mole of MnO₂

                                 OR

4x36.5 g of HCl reacts with 1x87g of MnO₂

48.5 g of HCl reacts with M of MnO₂

M = ( 48.5 x 1 x 87 ) / ( 4x36.5)

    = 28.9 g of MnO₂

So 34.3 - 28.9 = 5.4 g of MnO₂ left unreacted so it is the excess reactant.

Since all the mass of HCl completly reacted , HCl is the limiting reactant.

From the balanced Equation , 4 moles of HCl produces 1 mole of Cl₂

                                                                            OR

                                               4x36.5 g of HCl produces 71 g of Cl₂

                                                 48.5 g of HCl produces N g of Cl₂

                                                         N = ( 48.5 x 71 ) / ( 4x36.5)

                                                            = 23.6 g of Cl₂   -----> this is the theoretical yield

Give that the % yield = 77.9 %

% yield = ( actual yield / theoretical yield ) x 100

77.9   = ( actual yield / 23.6 ) x 100

actual yield = (77.9 x 23.6 ) / 100

                   = 18.4 g

Therefore the actual yield of Cl₂ is 18.4 g