Question

Magnesium oxide can be made by heating magnesium metal in the presence of gaseous oxygen. When 10.1 grams of magnesium is allowed to react with 10.5 grams of molecular oxygen, 11.9 grams of magnesium oxide is collected. What is the percent yield of this reaction?

Answer 1

the overall reaction:

2Mg + O₂ ----------> 2MgO

10.1g 10.5g 11.9g

Let's calculate the theorical moles:

moles Mg = 10.1 / 24 = 0.4208 moles

moles O2 = 10.5 / 32 = 0.3281 moles

Now let's calculate the limitant reactant:

2 mole Mg -------> 1 mole O2

0.4208 Mg --------> x

x = 1*0.4208 / 2 = 0.2104 moles O2

we have 0.3281 mole of O2, which mean that the Mg is the limitant reactant.

Now, these moles would be the moles produced of MgO, let's calculate the mass:

m MgO = 0.4208 mol * (24+16) g/mol = 16.832 g

Finally the %yield:

% = (11.90/16.832) * 100 = 70.7%

Hope this helps