Question

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

2Mg(s)+O2(g)→2MgO(s)

When 10.0 g Mg is allowed to react with 10.5 g O2, 11.9 g MgO is collected.

1. Determine the theoretical yield for the reaction. (Express your answer in grams)

2. Determine the percent yield for the reaction. (Express your answer as a percent.)

Answer 1

2Mg(s)+O2(g)----> 2MgO(s)

2 moles Mg needs 1 mole O2 to yield 2 mole MgO.

mass of Mg present = 10 g
molar mass of Mg = 24.31 g/mol
therefore number of moles of Mg Present = mass of Mg/molar mass of mg = 10g/24.31 g
= 0.411 mols

Also mass of O2 present = 10.5 g
molar mass of O2 = 32 g/mol
therfore number of moles of O2 present = mass of O2/molar mass of O2
= 10.5 g/32 g/mol= 0.328 mols

Since 0.328 mols O2 needs 2*0.328 mols Mg [since their mol ratio = 1:2]
=0.656 mols of Mg is needed
But only 0.411 mols Mg is present. Thus Mg will be the limiting reactant.
Mg will consume completely on reaction with O2 .
Moles of Mg present will decide amount of product formed

2moles Mg produce 2 moles MgO,[their mol ratio =1:1]

0.411 mol Mg will produce 0.411 mol MgO

Thus number of moles of MgO formed = 0.411 mols
we have molar mass of MgO =40.304 g/mol

thus mass of MgO formed = number of Moles of MgO formed*molar mass of MgO
= 0.411 mols*40.304 g/mol = 16.56 g

Thus the theoretical yield of MgO =16.56 g

Actual yield = 11.9 g

Therefore % yield = Actual yield/theoretical yield *100 = 11.9/16.56 *100
= 71.86 %

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Thank you