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The theoretical yield of a reaction is 5.00 moles of potassium permanganate (KMnO4) .

If the reaction actually produces 652.5 g KMnO4 , what is the percent yield of the reaction?

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anfabba15

Answer:

The percent yield of the reaction is 82.5%

Explanation:

Let's work with moles to get the percent yield.

Mass / Molar mass =

652.5 g / 158.03 g/m = 4.13 moles

If the theoretical yield of the reaction is 5 moles but we only made 4.13 moles, the percent yield will be:

(Produced yield / Theoretical yield) . 100 =

(4.13 / 5) . 100 = 82.5 %

Answer:

The % yield is 82.58 %

Explanation:

Step 1: Data given

theoretical yield = 5.00 moles of KMnO4

Molar mass of KMnO4 = 158.03 g/mol

Mass of KMnO4 produced = 652.5 grams

Step 2: Calculate mass of 5.00 moles KMnO4

Mass = moles * molar mass

Mass KMnO4 = 5.00 moles * 158.03 g/mol

Mass KMnO4= 790.15 grams = theoretical yield

Step 3: Calculate % yield

%yield = (actual mass produced / theoretical mass )*100 %

% yield = (652.5 / 790.15)* 100 %

% yield = 82.58%

The % yield is 82.58 %

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