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Determine the percent yield for the reaction between 28.1 g of Sb4O6 and excess C if 17.3 g of Sb is recovered along with an unkown amount of CO

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greekfreekisdbz

The answer is 73.71%. Hope I helped plus mark as brailiest ^-^

pstnonsonjoku

The percent yield is 37%.

The equation of the reaction is; Sb4O6  + 6C ----> 4Sb + 6CO

Number of moles of Sb4O6  = 28.1 g/292 g/mol = 0.096 moles

We know from the question that C is the reactant in excess.

If 1 mole of Sb4O6 yields 4 moles of Sb

0.096 moles of Sb4O6 yields 0.096 moles × 4 moles/1 mole = 0.384 moles

Theoretical yield of Sb =  0.384 moles ×  122 g/mol = 46.8 g

Percent yield = 17.3 g /46.8 g × 100/1 = 37%

Learn more about percent yield: https://brainly.com/question/731147?

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