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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

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Answer:

[tex]Y_A=92.1\%\\\\Y_B=89.6\%[/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]

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

[tex]Y_A=\frac{83.67g}{90.9g} *100\%=92.1\%\\\\Y_B=\frac{81.35g}{90.9g} *100\%=89.6\%[/tex]

Regards!

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