Question

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

Answer 1

Answer:

$Y_A=92.1\%\\\\Y_B=89.6\%$

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:

$m_{KMnO_4}=50.0gMnO_2*\frac{1molMnO_2}{86.94gMnO_2}*\frac{2molKMnO_4}{2molMnO_2} *\frac{158.034gKMnO_4}{1molKMnO_4} \\\\m_{KMnO_4}=90.9gKMnO_4$

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

$Y_A=\frac{83.67g}{90.9g} *100\%=92.1\%\\\\Y_B=\frac{81.35g}{90.9g} *100\%=89.6\%$

Regards!