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 Purpose First step is

We have to know the theoretical yield KMnO₄

Potassium 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 theoretical 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%

i wish i could but this one is weird give me abc and i can solve it.