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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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sebassandin

Answer:

[tex]m_{KMnO_4}=90.9gKMnO_4[/tex]

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:

[tex]m_{KMnO_4}=50.0gMnO_2*\frac{1molMnO_2}{86.94gMnO_2}*\frac{2molKMnO_4}{2molMnO_2} *\frac{158.034gKMnO_4}{1molKMnO_4} \\\\m_{KMnO_4}=90.9gKMnO_4[/tex]

Regards!

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