Question

#1: The theoretical yield of a reaction is 5.00 moles of potassium permanganate (KMnO4). If the reaction actually produces 695.2 g KMnO4, what is the percent yield of the reaction?

A. 78%

B. 83%

C. 88%

D. 120%

Answer 1

To calculate the percent yield, the actual yield in moles of potassium permanganate was found to be 4.40 moles from 695.2 g, using the molar mass of KMnO4. Then, the percent yield was calculated using the theoretical yield of 5.00 moles, resulting in 88%.

Therefore, the correct answer is C. 88%

Step-by-step explanation:

To calculate the percent yield of the reaction, we first need to know the moles of potassium permanganate (KMnO4) that correspond to 695.2 g. We use the molar mass of KMnO4 which is 158.04 g/mol to convert grams to moles.

Actual yield in moles = mass (g) / molar mass (g/mol) = 695.2 g / 158.04 g/mol = 4.40 moles of KMnO4

Now, with the theoretical yield of 5.00 moles, we can find the percent yield using the formula:

Percent yield = (actual yield / theoretical yield) × 100%

Percent yield = (4.40 moles / 5.00 moles) × 100% = 88%

Therefore, the correct answer is C. 88%

Answer 2

The formula actual yield / theoretical yield is used to calculate the percent yield of a reaction. This value is a measure of how much product is produced relative to the what is supposed to be produced. We calculate as follows:

Percent yield = actual yield / theoretical yield x 100
Percent yield = 695.2 g ( 1 mol / 158.034 g) / 5 x 100
Percent yield = 88% <------OPTION C