2 Mg(s) + O_{2}(g) → 2 MgO(s)

Calculate the amount of MgO produced from each reactant and determine the limiting reactant:

mass Mg (molar mass) → moles Mg (mole-to-mole comparison) → moles MgO (molar mass) → mass MgO

molar mass Mg = 24.305 g/mol

molar mass MgO = 40.305 g/mol

$\mathbf{mass}\mathbf{}\mathbf{MgO}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{1}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Mg}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Mg}}}{\mathbf{24}\mathbf{.}\mathbf{305}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Mg}}}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{MgO}}}{\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Mg}}}\mathbf{\times}\frac{\mathbf{40}\mathbf{.}\mathbf{305}\mathbf{}\mathbf{g}\mathbf{}\mathbf{MgO}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{MgO}}}$

**mass MgO = 16.75 g MgO**

mass O_{2} (molar mass) → moles O_{2} (mole-to-mole comparison) → moles MgO (molar mass) → mass MgO

Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is

2 Mg(s) + O_{2}(g) → 2 MgO(s)

When 10.1 g Mg is allowed to react with 10.5 g O_{2}, 12.0 g MgO is collected.

Determine the theoretical yield for the reaction.

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