Question

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 (MnO2).

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

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

2 MnO2 + 2 KOH + O2 → 2 KMnO4 + H2

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Answer 1

`m_(KMnO_4)=90.9gKMnO_4`

Step-by-step explanation:

Hello there!

In this case, according to the given chemical equation for the reaction for the production of potassium permanganate, we can see a 2:2 mole ratio of this product to the starting manganese (II) oxide, which means, we can calculate the theoretical yield of the former via stoichiometry:

`m_(KMnO_4)=50.0gMnO_2*(1molMnO_2)/(86.94gMnO_2)*(2molKMnO_4)/(2molMnO_2) *(158.034gKMnO_4)/(1molKMnO_4) \\\\m_(KMnO_4)=90.9gKMnO_4`

Regards!