Question

Given the chemical equation, 2Mg + O2 —> 2MgO, when 2.2 g Mg react with 3.6 g of O2, 2.7 g MgO were obtained. What is the percent yield in the reaction?

Answer 1

Always convert to moles when comparing compounds.
Molar mass of a compound is the sum of it's atomic molar mass units.

Mg = 24.3 g/mol Mg
O2 = 16 + 16 = 32 g/mol O2
MgO = 24.3 + 16 = 40.3 g/mol MgO

Determine the moles of each reactant/product.
2.2 g Mg * (1 mol/24.3 g Mg) = 0.09 mol Mg
3.6 O2 * (1 mol/32 g O2) = 0.1125 mol O2
2.7 g MgO * (1 mol/40.3 g MgO) = 0.067 mol MgO
 
Check if there's a limiting reagent. For every 1 O2 we need 2 Mg
0.1125 mol O2 * 2 = 0.225 mol Mg needed.
So Mg is a limiting reagent. We have plenty of O2 which is typically the case when oxygen is a reactant.

Figure out how much product should form based on the moles of limiting reagent. For every 2 Mg 2 MgO are formed. So it's a 1:1 ratio.
0.09 mol Mg ---> 0.09 mol MgO

compare the expected 0.09 mol MgO to the actual 0.067 mol MgO obtained. Calculate the percent obtained.
(0.067 mol MgO obtained) / (0.09 mol MgO expected) * 100 = 74.44 % yield