Final answer:
Magnesium (Mg) is the limiting reactant in the reaction with oxygen (O2) to form magnesium oxide (MgO), as the moles of Mg present are less than the stoichiometrically required moles to fully react with the available oxygen.
Step-by-step explanation:
To determine the limiting reactant in a chemical reaction between magnesium and oxygen, we compare the moles of each reactant to their stoichiometric coefficients in the balanced equation. The balanced equation for the reaction between magnesium and oxygen to form magnesium oxide is:
2 Mg + O2 → 2 MgO
The molar mass of magnesium (Mg) is 24.31 g/mol, and for oxygen gas (O2) it is 32.00 g/mol. To find the moles of each reactant:
- Divide the mass of Mg by its molar mass to find the moles of Mg: 4.210 g Mg / 24.31 g/mol = 0.1732 mol Mg.
- Divide the mass of O2 by its molar mass to find the moles of O2: 4.130 g O2 / 32.00 g/mol = 0.1291 mol O2.
According to the stoichiometry of the balanced reaction, magnesium reacts with oxygen in a 2:1 mole ratio. Therefore, we need twice as many moles of magnesium as we do of oxygen.
To determine how many moles of Mg are needed to react with the O2 present, we multiply the moles of O2 by 2: 0.1291 mol O2 x 2 = 0.2582 mol Mg required. But we only have 0.1732 mol Mg, which is less than the required amount, making magnesium the limiting reactant.
The amount of MgO that can be produced is limited by the moles of the limiting reactant Mg, which is 0.1732 mol Mg times the molar mass of MgO (40.31 g/mol), yielding approximately 3.98 g MgO as the theoretical yield.