Answered

Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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?

You must show all work to receive full credit.

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

2 MnO2 + 2 KOH + O2 → 2 KMnO4 + H2

Sagot :

sebassandin

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!

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.