10.8 g of magnesium is combined with 10.0 liters of oxygen to form magnesium oxide. Calculate the mass (in grams) of magnesium oxide that will be produced. 2Mg + O2 2MgO

Md Ariful Islam

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Solution:

The balanced chemical equation:

2Mg + O2 → 2MgO

According to the chemical equation: n(Mg)/2 = n(O2) = n(MgO)/2

Molar mass of Mg is 24 g/mol.

Moles of Mg = 10.8 g × (1 mol / 24 g) = 0.45 mol

Molar volume of O2 at STP is 22.4 L/mol.

Moles of O2 = 10.0 L × (1 mol / 22.4 L) = 0.446 mol

Divide by coefficients of balanced equation:

magnesium (Mg) ⇒ 0.45 mol / 2 mol = 0.225

oxygen (O2) ⇒ 0.446 mol / 1 mol = 0.446

Magnesium is the lower value. It is the limiting reagent.

Therefore,

n(MgO) = n(Mg) = 0.45 mol.

Molar mass of MgO is 40 g/mol.

Mass of MgO = Moles of MgO × Molar mass of MgO;

m(MgO) = (0.45 g) × (40 g/mol) = 18.0 g.

m(MgO) = 18.0 g

Answer: 18.0 grams of magnesium oxide (MgO) will be produced.

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