4HCl(aq)+MnO2(s) ⟶ MnCl2(aq)+2H2O(l)+Cl2(g) A sample of 41.9g MnO2 is added to a solution containing 44.3 g HCl .

Question

What is the limiting reactant?

HCl

or

MnO2

What is the theoretical yield of Cl2?

If the yield of the reaction is 

 what is the actual yield of chlorine?

Answer 1

4HCl(aq) + MnO₂(s) ==> MnCl₂(aq) + 2H₂O(l) + Cl₂(g) ... balanced equation (re-dox reaction)

moles HCl present = 44.3 g x 1 mol/36.46 g = 1.215 moles HCl

moles MnO₂ present = 41.9 g x 1 mol/86.94 g = 0.4819 moles MnO₂

Since the mol ratio in the balanced equation is 4 HCl to 1 MnO₂, the HCl is LIMITING.

Theoretical yield of Cl₂ = 1.215 mol HCl x 1 mol Cl₂/4 mol HCl x 70.9 g Cl₂/mol = 21.5 g Cl₂ theoretical yield

Actual yield = 21.5 g x 0.833 = 17.9 g Cl₂ actual yield

Answer 2

If you perform two gram to gram calculations, you can determine both the amount of Cl₂ produced AND the limiting reactant.

Grams of substance A must be converted to moles of substance A. Moles of substance A must be converted to moles of substance B. Moles of substance B must be converted to grams of substance B.

1st calculation: assume that HCl runs out first (is the limiting reactant) and MnO2 is in excess.

44.3 g HCl • (1 mol HCl / 36.46 g HCl) • (1 mol Cl₂ / 4 mol HCl) • (70.9 grams Cl₂ / 1 mol Cl₂) = 21.5 g Cl₂ produced IF HCl is limiting.

2nd calculation: assume that MnO₂ runs out first (is the limiting reactant) and HCl is in excess.

41.9 g MnO₂ • (1 mol MnO₂ / 86.94 g MnO₂) • (1 mol Cl₂ / 1 mol MnO₂) • (70.9 grams Cl₂ / 1 mol Cl₂) = 34.2 g Cl₂ produced IF MnO₂ is limiting.

By analyzing the amount PRODUCED for each calculation, we assume the reaction goes to completion. Therefore, the smaller amount of product is the true theoretical yield. 21.5 g Cl₂ is made, which means HCl must be limiting.

Percent yield is the (actual yield / theoretical yield) * 100 to get a percentage. (83.3/100) = x / 21.5 grams Cl2 produced. (0.833 * 21.5 = 17.9 g Cl₂ recovered in lab)