4.0 g Mg and 4.0 g O2 are placed in a container and magnesium oxide, MgO, forms. The Mg is totally consumed but 1.4 g O2 remains. How much magnesium oxide formed

The correct answer is: 6.6 g MgO

Explanation:

First we have to write and balance the chemical reaction as follows:

2Mg(s) + O₂(g) → 2MgO(s)

That means that 2 moles of Mg(s) react with 1 mol of O₂(g) to give 2 moles of MgO(s). If Mg is totally consumed and a mass of O₂ remains unaltered after reaction, the limiting reactant is Mg. We use the limiting reactant to calculate the mass of product.

According to the balanced chemical equation, 2 moles of Mg(s) produce 2 moles of MgO(s).

2 moles Mg = 2 mol x molar mas Mg= 2 mol  x 24.3 g/mol = 48.6 g Mg

2 moles MgO= 2 mol x (molar mass Mg + molar mass O) = 2 mol x (24.3 g/mol + 16 g/mol) = 80.6 g MgO

The stoichiometric ratio is 80.6 g MgO/48.6 g Mg. So, we multiply this ratio by the mass of consumed Mg (4.0 g) in order to obtain the produced mass of MgO:

4.0 g Mg x 80.6 g MgO/48.6 g Mg = 6.63 g MgO

6.6 grams of magnesium oxide are formed.