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 10.15 g Mg is allowed to react with 10.40 g O2, 11.85 g MgO is collected.

Determine the theoretical yield for the reaction.

Express your answer in grams.

Determine the percent yield for the reaction.

Express your answer as a percent.

Answer 1

Atomic weighs : Mg =24 O =16

Molecular weights : MgO= 24+16=40, O2=32

Moles : Mass/ Molecular weight

Moles : Mg= 10.15/24=0.4229 O2= 10.4/32= 0.325 MgO= 11.85/40=0.29625

From the reaction 2 Mg+O2----> 2MgO, two moles of Mg reacts with 1 mole of oxygen to give 2 moles of Mgo

stoichiometric ratio of Mg : O2= 2:1

Actual ratio of reactants taken = 0.4229 : 0.325 dividing bt 0.325

the ratio become 0.4229/0.325 :1 =1.3 :1

So Mg is the limting reactant since requires 2 for one mole of oxygen where as supplie is only 1.3

Limiting reactants govern the exent of reaction and this 10.15 gms

from the reaction 2 mole of Mg= 48 gms

48 gms of Mg gives rise to 2*40=80 gms of MgO

10.15 gm of Mg gives rise to 80*10.15/48 gms of MgO=16.91 gms of MgO

Theoretical yiled is 16.91 gm

Actual yiled is 11.85 gms

percent yiled= 100* actual yiled/ theorerical yiled= 100*11.85/16.91=70.07%