Question

Part A You may want to reference (Pages 299 306) Section 7.5 while completing this problem. Determine the limiting reactant for the reaction. Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is 2 Mg(s)+O2(g)2MgO(s) Mg(s) O2(g) When 10.1 g Mg is allowed to react with 10.5 g O2, 11.9 g MgO is collected Submit Request Answer Part B reaction the theoretical yield Determi Express your answer in grams. Tm Request Answer Submit Part C Determine the percent yield for the reaction Express your answer as a percent να ΑΣφ 5 Submit Request Answer

Answer 1

A) Equation: 2Mg +0, 2Mgo No of moles of MgO produced from Mg = 10.1gx 1 mol Mg 2 mol Mgo 24.31 g 2 mol Mg = 0.415 mol Mgo No of moles of MgO produced from 0 = 10.5 8x1 mol O 2 mol Mgo 32.0 g 1 mol Oy = 0.656 mol Mgo Since Mg imits the production of MgO, Mg acts as the limiting reactant 40.31 g b) Amount of MgO produced=0.415 mol MgOx- 1 mol Mgo = 16.7 g c) Actual yield of MgO = 11.9 g Theoretical yield of MgO = 16.7 g Actual yield Percent yield = -X100 Theoretical yield 11.98 x100 16.7 g = 71.2% 71%