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
