During a laboratory experiment, 9.68 g of iron reacted with excess sulfur to form 13.8 grams of iron II sulfide. Calculate the percent yield of this experiment. What would be the percentage yield if the Theoretical yield was 15.2 grams?

Percent yield = 90.8%

Explanation:

The reaction of Fe with S to produce FeS is:

Fe + S → FeS

Where the moles of Fe added in excess of S are the moles of FeS

Now, percent yield is defined as 100 times the ratio between actual yield (13.8g of FeS) and theoretical yield (15.2g FeS):

Percent yield = 13.8g / 15.2g * 100

the answer is d : ) have a good da