If 21.42g of kmno4 is actually produced what is the percent yield

Percentage yield  = 6.776%

Explanation:

Data Given:

Actual yield of KMnO₄ = 21.42g

Percentage Yield = ?

Formula Used to find Percent yield

                 Percentage yield = Actual yield/ theoretical yield x 100        (1)

For this Pupose First step is

We have to know the theoretical yield KMnO₄

Potasium permagnate form from MnO₄ and KOH in the presence of Oxygen by heating, in 1st step in second 2K₂MnO₄ react with HCl and give KMnO₄ .

The Reactions of formation of KMnO₄

1st Step                      

2MnO₄ + 4KOH + O₂  > 2K₂MnO₄ + 2H₂O

2nd Step

2K₂MnO₄ + 4HCl >2 KMnO₄ + H₂O + 4KCl

So form the above equation we come to know it produced 2 Mole of KMnO₄

Now we will calculate the mass of KMnO₄ by mass formulae

     mass of KMnO₄ = Number of mole of KMnO₄ x Molar Mass of KMnO₄

Molar Mass of KMnO₄ = 158.034g /mol

Put value in Mass Formula

 mass of KMnO₄ = 2 mol x  158.034g/mol

mass of KMnO₄ = 316.1 g

So the theoratical yield per standard reaction = 316.1 g

Now put all values in equation 1

           Percentage yield = Actual yield/ theoretical yield x 100  

           Percentage yield  = 21.42g / 316.1 g x 100

          Percentage yield =  0.0678 x 100

          Percentage yield  = 6.776%

the answer is d. 74.02