Question

mass of magnesium oxide produced when 9.25 grams of magnesium reacts with 7.23 grams of oxygen.Identify the limiting reactant

Answer 1

The limiting reactant is magnesium, and the excess reactant is oxygen. The reaction produces 15.33 grams of magnesium oxide when 9.25 grams of magnesium reacts with 7.23 grams of oxygen.

Step-by-step explanation:

To identify the limiting reactant in the reaction between magnesium and oxygen to form magnesium oxide (MgO), we need to follow these steps:

1. Write the balanced chemical equation for the reaction:
   2 Mg(s) + O2(g) → 2 MgO(s).
2. Calculate the number of moles of each reactant present. The molar mass of Mg is 24.31 g/mol and O2 is 32.00 g/mol.
3. Using the given masses, calculate moles of Mg: 9.25 g Mg × (1 mol Mg/24.31 g Mg) = 0.3804 moles of Mg.
   Calculate moles of O2: 7.23 g O2 × (1 mol O2/32.00 g O2) = 0.2259 moles of O2.
4. Compare the stoichiometric ratio from the balanced equation with the actual mole ratio to determine which reactant is limiting. According to the balanced equation, the ratio is 2 moles of Mg to 1 mole of O2. We have 0.3804 moles of Mg, which requires 0.1902 moles of O2 to completely react (since 2:1 ratio).
5. Since 0.2259 moles of O2 are available and only 0.1902 moles are required, magnesium is the limiting reactant, and oxygen is in excess.
6. To calculate the mass of magnesium oxide produced, use the moles of the limiting reactant (Mg) and the molar ratio from the balanced equation. Since 1 mole of Mg produces 1 mole of MgO, 0.3804 moles of Mg will produce 0.3804 moles of MgO.
   The molar mass of MgO is 40.31 g/mol. Therefore, 0.3804 moles × 40.31 g/mol = 15.33 g of MgO are produced.