Question

Scientist A produces 83.67 g KMnO4 while Scientist B produces 81.35 g KMnO4.

What is the percent yield for Scientist A?

What is the percent yield for Scientist B?

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

2 MnO2 + 2 KOH + O2 → 2 KMnO4 + H2

Answer 1

`Y_A=92.1\%\\\\Y_B=89.6\%`

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`

Now, we are able to compute the percent yields, by using the actual yield each scientist got:

`Y_A=(83.67g)/(90.9g) *100\%=92.1\%\\\\Y_B=(81.35g)/(90.9g) *100\%=89.6\%`

Regards!