Answer : The theoretical yield and the percent yield is, 27.1 grams and 80.4 % respectively.
Explanation:
First we have to calculate the moles of KO₂ and CO₂.
and,
As we know that, 1 mole of substance occupies 22.4 L volume of gas.
As, 22.4 L volume of CO₂ present in 1 mole of CO₂ gas
So, 29.0 L volume of CO₂ present in mole of CO₂ gas.
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
From the balanced reaction we conclude that
As, 4 mole of react with 2 mole of
So, 0.392 moles of react with moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 4 mole of react to give 2 mole of
So, 0.392 moles of react to give moles of
Now we have to calculate the mass of
Molar mass of = 110.98 g/mole
Now we have to calculate the percent yield of the reaction.
Experimental yield = 21.8 g
Theoretical yield = 27.1 g
Now put all the given values in this formula, we get:
Therefore, the percent yield of the reaction is, 80.4 %