Determine the percent yield for the reaction between 162 g of Sb2S3 and excess oxygen if 55 g of Sb4O6 is recovered along with an unknown amount of sulfur dioxide gas?
2Sb2S3 + 9O2 --->Sb4O6 + 6SO2
How many mols Sb2S3 is 162 grams. That's 162/679.4 = about 0.24 but that's just an estimate. You should calculate more accurately.
How many mols Sb4O6 will that produce? That's about 0.24/2 = about 0.12
How many grams is 0.12 mols Sb4O6? That's
grams = mols x molar mass = 0.12 x 583 g/mol = about 68 g
%yield = (actual yield/theoretical yield)*100 = approx (55/68)*100 = ?
Post your work if you get stuck. Don't forget to redo ALL of the calculations because I've estimated. The answer is close but not correct. Also check those molar mass numbers.