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 42.1 g MnO2

is added to a solution containing 45.9 g HCl.

What is the limiting reactant?

A.MnO2

B.HCl

What is the theoretical yield of Cl2?

theoretical yield:_____?______g Cl2

If the yield of the reaction is 71.3%,

what is the actual yield of chlorine?

actual yield:____?_____g Cl2

Answer 1

Nicole,

The theoretical yield is based on the limiting reactant so once you've found it you can get to the yield.

The limiting reactant is the reactant that is used up completely in the reaction. It's the one that's done first. What this means is that the amount of product you get is based on how much of it is present. Once it's done, the reaction stops and that's as much product you'll get.

To figure out the limiting reagent, calculate the number of moles of each reactant available.

We have to convert our masses to moles as the balanced equation is the only thing that connects all these chemicals, and the balanced equation shows mole relationships.

Convert mass to moles by using the molar mass of each reactant:

For MnO₂ with a molar mass of 86.94 g/1 mol -------- 42.1 g x 1 mol/86.94 g = 0.484 mols MnO₂

For HCl with a molar mass of 36.46 g/1 mol ----------- 45.9 g x 1 mol/36.46 g = 1.26 mols HCl

So for this reaction we have 0.484 mols of MnO₂ and 1.26 mols of HCl

Based on the balanced equation we need 4 times as much HCl than MnO₂. When we compare 1.26 mols/0.484 mols we get 2.6. This shows that the moles of HCl are not 4 times as much as that of MnO₂. This means that the HCl is the limiting reagent as we do not have enough of it. This also means that it will determine how much Cl₂ is produced based on its ratio to Cl₂ in the balanced equation.

The balanced equation shows that for every 4 mols of HCl, 1 mole of Cl₂ is produced (a quarter). So, since we have 1.26 mols of HCl from the calculation above, 1/4 of that will be produced as Cl₂.

1.26 mols HCl x 1 mol Cl₂/4 mols HCl = 0.315 mols Cl₂

To convert this mass to g we multiply by Cl₂'s molar mass which is 70.906 g/1 mol

0.315 mols x 70.906 g/mol = 22.3 g Cl₂

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

71.3 = (actual yield/22.3 g) x 100

0.713 = actual yield/22.3 g

0.713 x 22.3 g = actual yield

15.9 g = actual yield