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Consider the following reaction:
2H2S (g) + 3O2 (g) --> 2SO2 (g) + 2H2O (g)
If O2 was the excess reagent, 1.3 moles of H2S were consumed, and 18.7 g of wat were collected after the reaction has gone to completion, what is the percent yield of the equation?
What is the theoretical yield?
1.3 moles H2S x (2 moles H2O/2 moles H2S) = 1.3 moles x (1/1) = 1.3 moles H2O produced. g H2O = moles x molar mass = 1.3 moles x 18 g/mol = about 23 estimated theoretical yield. Therefore, percent yield = (18.7/theoretical yield)*100 = ?