If 0.66 g magnesium reacts exactly with 0.43 g oxygen, what mass of magnesium oxide will be produced?

Balanced equation for the reaction is;
2Mg + O₂ --> 2MgO
stoichiometry of Mg to O₂ is 2:1
we need to first find the limiting reactant 
number of Mg moles - 0.66 g / 23 g/mol = 0.029 mol
number of O₂ moles - 0.43 g / 32 g/mol = 0.013 mol
If Mg is the limiting reactant 
2 mol of Mg reacts with 1 mol of O₂
then 0.029 mol of Mg reacts with - 0.029/2 = 0.0145 mol 
but only 0.013 mol of O₂ is present
therefore O₂ is limiting reactant. amount of product formed depends on amount of limiting reactant present,
stoichiometry of O₂ to MgO is 1:2
when 0.013 mol of O₂ reacts - 0.026 mol of MgO is formed 
mass of MgO produced is - 0.026 mol x 39 g/mol = 1.014 g
1.014 g of MgO formed

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer