Now, the information is not complete as there should be the mass of at least one substrate in the question,
So let's assume the mass of Potassium permanganate is 158g
From the reaction,
1mole of "KMnO_4" produces 1 mole of "KC_7H_5O_2"
158g of KMnO4 = 1 mole of KMnO4.
Therefore 1 mole of potassium benzoate is produced.
1 mole of potassium benzoate = 160g.
Therefore, theoretical yield = 160g
Actual yield = Theoretical yield/100% × %yield
= 160g/100% × 83%
= 132.8g
Therefore, when 158g of potassium permanganate is used in the reaction, 132.8g of potassium benzoate is produced.
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