Magnesium oxide was obtained by reacting 12 g of magnesium with 12 g of oxygen.

Magnesium oxide was obtained by reacting 12 g of magnesium with 12 g of oxygen. Determine its mass. Which of the starting substances and in what quantity will remain in excess?

To solve the problem, let’s compose the reaction equation:
2Mg + O2 = 2MgO – reaction of the compound, magnesium oxide was obtained;
Let’s calculate the molar masses of substances:
M (Mg) = 24.3 g / mol;
M (O2) = 32 g / mol;
M (MgO) = 24.3 + 16 = 40.3 g / mol;
Determine the amount of moles of magnesium, oxygen:
Y (Mg) = m / M = 12 / 24.3 = 0.49 mol (substance in excess);
Y (O2) = m / M = 12/32 = 0.37 mol (deficient substance);
Thus, the calculations are carried out for the substance in deficiency.
Let’s make the proportion:
X mol (MgO) – 0.37 mol (O2);
-2 mol – 1 mol hence, X mol (MgO) = 2 * 0.37 / 1 = 0.74 mol;
We calculate the mass of magnesium oxide by the formula:
m (MgO) = Y * M = 0.74 * 40.3 = 29.82 g.
Answer: during the reaction, magnesium oxide with a mass of 29.82 g is released.