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

Question

10.) 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  is added to a solution containing,What is the theoretical yield of Cl2?

If the yield of the reaction is 

 what is the actual yield of chlorine?

Answer 1

I think your question got partially cut off. I googled what you wrote, however and I think I understand what the question was.

I'm guessing you were given grams of MnO₂ and HCl

We will use stoichiometry to solve this problem.

Use the periodic table to find the molar masses of each element used in the reaction:

Mn = 54.938 grams/mole

O = 15.999 grams/mole

H = 1.008 grams/mole

Cl = 35.45 grams/mole

MnO₂ has a mass of (54.938 + 2(15.999)) = 86.9368 g/mol

HCl has a mass of (1.008 + 35.45) = 36.458 g/mol

Now we need to find the limiting factor. I don't know what the mass of each substance was that you were given, so I will substitute with words.

given grams MnO₂(s) x 1 mol MnO₂ = [given grams / 86.9368] mols of MnO₂(s)

86.9368 grams

given grams HCl(l) x 1 mol HCl = [given grams / 36.458] mols of HCl(l)

36.458

You will have the initial amounts of both substances to put in the numerator for "given grams".

Once you calculate those values, divide the moles calculated to find the molar ratio of HCl and MnO₂

mols of MnO₂(s) / mols of HCl(l)

If the answer to this fraction, for example, is 3, then there should be 3x the amount of MnO₂ than HCl.

Now we look to our original equation:

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

The coefficients tell us the ideal ratio for the reaction to fully complete.

We need 4 HCl molecules for every 1 MnO₂ molecule, so the ideal ratio of HCl:MnO₂ is 4:1

To determine the limiting factor, we compare the two ratios we just found.

If the molar ratio is less than the ideal 4 HCl to every 1 MnO₂ , then HCl is the limiting factor. If there is 4x the amount of HCl or higher in the molar ratio, then MnO₂ is the limiting factor.

Since I can't see what your actual yield was, I can't help much with that part of the question. I did find a really good article to help you out when you come back to this problem.

https://www.wikihow.com/Calculate-Theoretical-Yield