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 41.5 g MnO2 is added to a solution containing 48.7 g HCl.

What is the theoretical yield of Cl2?

theoretical yield:___gCl2

If the yield of the reaction is 76.3%, what is the actual yield of chlorine?

actual yield:___g Cl2

Answer 1

Number of moles of A=( mass of A)/(molecular mass of A)

Number of moles of MnO2 = (41.5g)/(87g/mol)

Number of moles of Mno2 = 0.477 mol

Number of moles ofHCl = (48.7g/35.5g/mol)

Number of moles of HCl = 1.3718mol

dividing by coefficient of reactant

for MnO2 = (0.477)/1 = 0.447

for HCl = (1.3718)/4 = 0.3429

so HCl is a limiting reagent

according to HCl and stoichiometry of reaction

4 mole HCl react with 1 mole of MnO2 and form 1 mole Cl2

So 1 mole HCl react with 1/4 mole Mno2 and form 1/4 mole Cl2

Hence 1.3718 mole of Hcl react with 1×1.3718/4 mole of MnO2 and form 1×1.3718/4 mole of Cl2

Number of moles of Cl2 = 1×1.3718/4 mole

Number of moles of Cl2 = 0.3429 mol

Mass of Cl2 = number of moles of Cl2 × molecules mass of Cl2

Mass of Cl2 = (0.3718 mole)(71g/mole)

Mass of Cl2 = 24.350 g ( theoritical yield)

%yield = (yield)100/theoritical yield

76.3= (yield)100/24.350g

Yield = (76.3)(24.350g)/100

Yield = 18.579g