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.
![\text{Moles of }KO_2=\frac{\text{Mass of }KO_2}{\text{Molar mass of }KO_2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DKO_2%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DKO_2%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DKO_2%7D)
![\text{Moles of }KO_2=\frac{27.9g}{71.10g/mol}=0.392mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DKO_2%3D%5Cfrac%7B27.9g%7D%7B71.10g%2Fmol%7D%3D0.392mol)
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:
![4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2](https://tex.z-dn.net/?f=4KO_2%2B2CO_2%5Crightarrow%202K_2CO_3%2B3O_2)
From the balanced reaction we conclude that
As, 4 mole of
react with 2 mole of ![CO_2](https://tex.z-dn.net/?f=CO_2)
So, 0.392 moles of
react with
moles of ![CO_2](https://tex.z-dn.net/?f=CO_2)
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 ![K_2CO_3](https://tex.z-dn.net/?f=K_2CO_3)
From the reaction, we conclude that
As, 4 mole of
react to give 2 mole of ![K_2CO_3](https://tex.z-dn.net/?f=K_2CO_3)
So, 0.392 moles of
react to give
moles of ![K_2CO_3](https://tex.z-dn.net/?f=K_2CO_3)
Now we have to calculate the mass of ![K_2CO_3](https://tex.z-dn.net/?f=K_2CO_3)
![\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DK_2CO_3%3D%5Ctext%7B%20Moles%20of%20%7DK_2CO_3%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DK_2CO_3)
Molar mass of
= 110.98 g/mole
![\text{ Mass of }K_2CO_3=(0.196moles)\times (138.21g/mole)=27.1g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DK_2CO_3%3D%280.196moles%29%5Ctimes%20%28138.21g%2Fmole%29%3D27.1g)
Now we have to calculate the percent yield of the reaction.
![\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100](https://tex.z-dn.net/?f=%5Ctext%7BPercent%20yield%7D%3D%5Cfrac%7B%5Ctext%7BExperimental%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100)
Experimental yield = 21.8 g
Theoretical yield = 27.1 g
Now put all the given values in this formula, we get:
![\text{Percent yield}=\frac{21.8g}{27.1g}\times 100=80.4\%](https://tex.z-dn.net/?f=%5Ctext%7BPercent%20yield%7D%3D%5Cfrac%7B21.8g%7D%7B27.1g%7D%5Ctimes%20100%3D80.4%5C%25)
Therefore, the percent yield of the reaction is, 80.4 %