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Determine the percent yield for the reaction between 98.7 g of
Sb2S3 and excess oxygen gas if 72.4 g of Sb4O6 is recovered along with an unknown amount of sulfur dioxide.

Sagot :

Lucillekayidn
Convert 98.7 g Sb2S3 to mols.
Convert mols Sb2S3 to mols Sb4O6 using the coefficients in the balanced equation.
Convert mols Sb4O6 to grams Sb4O6. (This is the theoretical yield.)
Convert grams Sb4O6 to percent Sb4O6.
%Sb4O6 = [72.4/theoretical yield]*100


So basically your answer is 85.5%
Oseniidn

The percent yield of the reaction would be 85.64%

Percent Yield

From the balanced equation of the reaction:

[tex]9 O_2 + 2 Sb_2S_3 ---> 6 SO_2 + Sb_4O_6[/tex]

Mole ratio of Sb2S3 to Sb4O6 is 2:1

Mole of 98.7g Sb2S3 = 98.7/339.7

                                   = 0.29 moles

Equivalent mole of Sb4O6 = 0.29/2

                                            = 0.145 moles

Mass of 0.145 Sb4O6 = 0.145 x 583

                                     = 84.535 g

Percent yield = yield/theoretical yield x 100

                       = 72.4/84.535 x 100

                        = 85.64%

More on percent yield can be found here: https://brainly.com/question/2506978?referrer=searchResults

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