Asked by jackson
Determine the percent yield for the reaction between 118 g of Sb2S3 and excess oxygen if 74 g of Sb4O6 is recovered along with an unknown amount of sulfur dioxide gas (include 2 decimal places in your answer).
___ Sb2S3 + ___ O2 ⟶ ___ Sb4O6 + ___ SO2
___ Sb2S3 + ___ O2 ⟶ ___ Sb4O6 + ___ SO2
Answers
Answered by
DrBob222
2Sb2S3 + 9O2 ⟶ Sb4O6 + 6SO2
mols Sb2S3 initially = grams/molar mass = ?
mols Sb4O6 formed = 1/2 mols Sb2S3 initially.
grams Sb4O6 = mols Sb4O6 x molar mass Sb4O6 = theoretical yield = TY
Actual yield from the problem = AY = 74 g.
% yield = (AY/TY)*100 ==> ?
Post your work if you get stuck.
mols Sb2S3 initially = grams/molar mass = ?
mols Sb4O6 formed = 1/2 mols Sb2S3 initially.
grams Sb4O6 = mols Sb4O6 x molar mass Sb4O6 = theoretical yield = TY
Actual yield from the problem = AY = 74 g.
% yield = (AY/TY)*100 ==> ?
Post your work if you get stuck.
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