Question

Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) arrow 2Hg (l) + O2 (g). If 4.00 moles of HgO decompose to form 1.50 moles of O2 and 603 g of Hg, what is the percent yield of this reaction?


26.7%

53.4%

75.0%

95.5%

Answer 1

4.00 moles of HgO ought to form
(4.00 mol) x (1 mol O2 / 2 mol HgO) = 2.00 mol of O2 in theory
(1.50 mol O2) / (2.00 mol ) = 0.750 = 75.0%