Answer:
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, t<em>he limiting reactant is Mg</em>. 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.
