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