1. Balanced Chemical Equation:
The reaction between magnesium (Mg) and oxygen (O₂) produces magnesium oxide (MgO):
2 Mg + O₂ → 2 MgO
2. Convert Masses to Moles:
* Mg:
* Moles of Mg = (mass of Mg) / (molar mass of Mg)
* Moles of Mg = (3.00 g) / (24.31 g/mol) = 0.123 mol
* O₂:
* Moles of O₂ = (mass of O₂) / (molar mass of O₂)
* Moles of O₂ = (2.20 g) / (32.00 g/mol) = 0.0688 mol
3. Determine the Limiting Reactant:
* Use the mole ratio from the balanced equation: The equation shows that 2 moles of Mg react with 1 mole of O₂.
* Calculate the moles of O₂ needed to react completely with the given moles of Mg:
* (0.123 mol Mg) * (1 mol O₂ / 2 mol Mg) = 0.0615 mol O₂
* Compare the calculated moles of O₂ needed with the actual moles of O₂ available:
* We need 0.0615 mol O₂, but we only have 0.0688 mol O₂. This means we have enough oxygen.
* Conclusion: Magnesium (Mg) is the limiting reactant because it will be completely consumed before all of the oxygen reacts.
4. Calculate the Theoretical Yield of MgO:
* Use the mole ratio from the balanced equation: The equation shows that 2 moles of Mg produce 2 moles of MgO.
* Calculate the moles of MgO produced from the limiting reactant (Mg):
* (0.123 mol Mg) * (2 mol MgO / 2 mol Mg) = 0.123 mol MgO
* Convert moles of MgO to grams:
* Mass of MgO = (moles of MgO) * (molar mass of MgO)
* Mass of MgO = (0.123 mol) * (40.30 g/mol) = 4.96 g MgO
Therefore:
* Limiting Reactant: Magnesium (Mg)
* Theoretical Yield of MgO: 4.96 g
