A company manufacturing KMnO4 wants to obtain the highest yield possible. Two of their research scientists are working on a technique to increase the yield. Both scientists started with 50.0 g of manganese oxide.

What is the theoretical yield of potassium permanganate when starting with 50.0 g MnO2?

The equation for the production of potassium permanganate is as follows:

2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2

The theoretical yield potassium permanganate, KMnO₄ when starting with 50.0 g MnO₂ is 90.8 g

Explanation:

Molar mass of MnO₂ = (55 + 2 × 16) = 87.0 g/mol

Molar mass of KMnO₄ = (39 + 55 + 4 × 16) = 158 g/mol

Moles of MnO₂ in 50 g = reacting mass / molar mass

where reacting mass = 50 g

Moles of MnO₂ in 50 g = 50 g /87 g/mol = 0.575 moles

The equation for the production of potassium permanganate is as follows:

2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2

From the equation of the reaction above, 2 moles of MnO₂ produces 2 moles of KmNO₄. The mole ratio of MnO₂ to KMnO₄ is 1 : 1

Therefore, 0.575 moles of MnO₂ will produce theoretically 0.575 moles of KMnO₄

Mass of 0.575 moles of KMnO₄ = number of moles × molar mass

Mass of 0.575 moles of KMnO₄ = 0.575 moles × 158 g/mol = 90.8 g of KMnO₄

Therefore, the theoretical yield potassium permanganate when starting with 50.0 g MnO₂ is 90.8 g