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

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 41.1 g MnO2 is added to a solution containing 49.7 g HCl.

a.) What is the limiting reactant?

b.) What is the theoretical yield of Cl2?

c.) If the yield of the reaction is 74.3%,74.3%, what is the actual yield of chlorine?

Answer 1

a) Using the given gram values, 41.1 g MnO₂ and 49.7 g HCl, convert each into moles by dividing by their respective molar masses (86.936 g/mol for MnO₂, and 36.461 g/mol for HCl):

41.1 g MnO₂ x (1 mol / 86.936 g) = 0.473 mol MnO₂

49.7 g HCl x (1 mol / 36.461 g) = 1.363 mol HCl

From here, we can determine which is the limiting reactant by converting each of them into one of the products given in the balanced chemical equation-- in this case, I am going to use stoichiometry to convert these molar values into moles of Cl₂, since we will be focusing on this product in later questions.

0.473 mol MnO₂ x (1 mol Cl₂/1 mol MnO₂) = 0.473 mol Cl₂

1.363 mol HCl x (1 mol Cl₂/4 mol HCl) = 0.341 mol Cl₂

Because of the fact that less Cl₂ is produced from HCl, that indicates that HCl is the limiting reactant.

b) The theoretical yield of Cl₂ (i.e., the maximum amount that can be theoretically produced) is dependent upon how much HCl was made from the limiting reactant. Using our answer from part a, we already determined that, because HCl is the limiting reactant, the amount of each product that is made depends on how much HCl is present. Once HCl runs out, the reaction stops.

We already know that 0.341 mol Cl₂ can be produced from the amount of HCl that we started with (49.7 g, which we then converted to 1.363 mol). The theoretical yield can be calculated by converting that mole value into grams:

0.341 mol Cl₂ x (70.906 g / 1 mol) = 24.18 g Cl₂ is the theoretical yield

c) We know that, of a possible 24.18 g Cl₂ that can be produced, only 74.3% of it was actually produced. Therefore, taking 74.3% of 24.18 g Cl₂:

0.743 x 24.18 g Cl₂ = 17.96 g Cl₂ is the actual yield