Magnesium oxide can be produced by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is:
2 Mg(s) + O2(g) → 2 MgO(s)
When 10.1 g of Mg react with 10.5 g of O2, 11.9 g of MgO are collected. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.
In this question the given mass of Mg = 10.1g
The mass of O2 = 10.5 g
Mass of MgO = 11.9 g
This is very simple, first do the theoretical yield since you need this number to get the percent yield.
We have to determine moles of reactants to see which (if any) is a limiting reagent. They stated it reacted with a certain amount of O2, they didn't say burned in air so you can't assume oxygen is present in excess!
The atomic weight of Mg = 24.305 g/mol and for O2 = 32.0g/mol
Hence number of mol of Mg = 10.1g/24.3g/mole =0.415 mole Mg
Similarly number of mole of O2 = 10.5g/32.0g/mole = 0.328 mole O2
From the balanced equation, you can see that 2 Mg's react with each O2 so 0.328 mole of O2 is more than enough to convert all the Mg to MgO; Mg is the limiting reagent.
Since the ratio of Mg reacted to MgO formed is 1 to 1 we would get 0.415 mole of MgO from 0.415 mole of Mg.
0.415mole [24.3 + 16.0]g/mole = 16.72g
So the theoretical yield (most MgO you could recover if reaction was 100% complete and no losses in recovering etc) is 16.72g