When 16.3 g of magnesium and 4.52 g of oxygen gas react, how many grams of magnesium oxide will be formed? identify the limiting and excess reactants.

22.77 g.

he limiting reactant is O₂, and the excess reactant is Mg.

Explanation:

Mg + 1/2O₂ → MgO,

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.

no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.

So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.

∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).

Using cross multiplication:

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.565 mol of Mg produce → 0.565 mol of MgO.

∴ The amount of MgO produced = no. of moles x molar mass = (0.565 mol)(40.3 g/mol) = 22.77 g.

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