Question

Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is:

2Mg(s)+O2(g)→2MgO(s) When 10.1 g of Mg are allowed to react with 10.5 g of O2, 12.4 g of MgO are collected.

Part A:   

Determine the limiting reactant for the reaction.

Express your answer as a chemical formula.

Part B: Determine the theoretical yield for the reaction.

Part C: Determine percent yield for the reaction.

Answer 1

2Mg(s)+O2(g)→2MgO(s)

no of moles of Mg   = W/G.A.Wt

                               = 10.1/24   = 0.42 moles

no of moles of O2   = W/G.M.Wt

                                = 10.5/32   = 0.328 moles

1 mole of O2 react with 2 moles of Mg

0.328 moles of O2 react with = 2*0.328/1   = 0.656 moles of Mg

Mg is limiting reactant

2 mole of Mg react with o2 to gives 2 moles of MgO

0.42 moles of Mg react with O2 to gives = 2*0.42/2   = 0.42 moles of MgO

mass of MgO   = no of moles * gram molar mass

                        = 0.42*40   = 16.8g

theoritical mass of MgO = 16.8g

percentage yield   = actual yield*100/theoritical yield

                              = 12.4*100/16.8   = 73.8% >>>>answer