Question

In a practical session, Hubert was asked to prepare potassium peroxide, K2O2 (s). He measured 3.910 g of potassium metal and burnt it in a container of oxygen gas with a mass of 2.560 g.

a) Write a chemical equation with state symbols for the reaction between potassium and oxygen.

b) Deduce which reactant, potassium or oxygen, is limiting.

c) Calculate the amount, in gram, of K2O2 (s) produced in the reaction.

d) It was found that the mass of K2O2 (s) obtained in the experiment was 3.609g. Calculate the percentage yield of K2O2 (s).

Answer 1

a-

The chemical reaction between Potassium (K) and Oxygen (O2) to form K2O2 is-

2K + O2 ---------> K2O2

i.e 2 mole of K need to react with 1 mole of O2 to produce 1 mole of K2O2

b-

Now given mass of K taken = 3.910 g

That means mols of K taken = mass / molar mass of K

= 3.910 g / 39.09 g/mole

= 0.1 moles

That means mols of O2 required for complete reaction of K = 0.1 moles/2

= 0.05 mols

Now given mass of O2 taken = 2.560 g

That means mols of O2 taken = mass / molar mass of O2

= 2.560 g / 32 g/mole

= 0.08 moles

That means here we have more than required mols of O2.

So the limiting reagent = K

And the excess reagent = O2

c-

So the complete reaction will be as per the limiting reagent i.e

0.1 mole K + 0.05 mole O2 ---------> 0.05 mole K2O2

i.e moles of K2O2 produced upon complete reaction = 0.05 mole

So mass of K2O2 produced upon complete reaction = moles * molar mass of K2O2

= 0.05 moles * 110.196 g/mol

= 5.51 g

d-

Now given actual mass of K2O2 produced = 3.609g

So % yeild = (theoritical yeild - experimental yeild) / theoritical yeild * 100

= (5.51 g - 3.609g) / 5.51 g * 100

= (1.901‬ g) / 5.51 g * 100

= 34.5 %