Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is: 2Mg(s) + O2(g) → 2MgO(s) When 10.0 g of Mg is allowed to react with 10.6 g of O2, 12.0 g of MgO is collected. a) Determine the limiting reactant for the reaction. b) Determine the theoretical yield for the reaction. c) Determine the percent yield for the reaction.
The Correct Answer and Explanation is:
Correct Answer:
Given:
- Mass of Mg = 10.0 g
- Mass of O₂ = 10.6 g
- Mass of MgO collected = 12.0 g
Balanced chemical equation:
2Mg(s) + O₂(g) → 2MgO(s)
Molar masses (g/mol):
Mg = 24.31
O₂ = 32.00
MgO = 40.30
a) Limiting Reactant
Step 1: Convert masses to moles
Moles of Mg = 10.0 g ÷ 24.31 g/mol ≈ 0.4113 mol
Moles of O₂ = 10.6 g ÷ 32.00 g/mol = 0.33125 mol
Step 2: Use mole ratio
From the balanced equation:
2 mol Mg reacts with 1 mol O₂
So 0.4113 mol Mg would require:
0.4113 ÷ 2 = 0.2057 mol O₂
Since we have 0.33125 mol O₂ (more than needed), Mg is the limiting reactant.
b) Theoretical Yield
Theoretical moles of MgO = moles of Mg used = 0.4113 mol (1:1 ratio with Mg)
Mass of MgO = 0.4113 mol × 40.30 g/mol = 16.57 g
c) Percent Yield
Percent yield = (actual yield ÷ theoretical yield) × 100
= (12.0 g ÷ 16.57 g) × 100 ≈ 72.4%
Explanation
When performing chemical reactions, it is essential to determine which reactant limits the amount of product that can be formed. This reactant is known as the limiting reactant. In this case, magnesium reacts with oxygen gas to form magnesium oxide. The chemical equation shows a 2:1 mole ratio between magnesium and oxygen. To identify the limiting reactant, we first convert the masses of each reactant to moles using their respective molar masses.
For magnesium, we divide 10.0 grams by its molar mass of 24.31 g/mol, giving about 0.411 moles. For oxygen gas, we divide 10.6 grams by 32.00 g/mol, yielding about 0.331 moles. Since magnesium requires only 0.5 moles of O₂ to fully react with 1 mole of Mg, the 0.411 mol of Mg would require about 0.206 mol of O₂, which we have in excess. Hence, magnesium is the limiting reactant.
Next, we determine the theoretical yield. Because the product (MgO) forms in a 1:1 molar ratio with magnesium, 0.411 moles of Mg should yield 0.411 moles of MgO. We convert this to grams using MgO’s molar mass (40.30 g/mol), resulting in a theoretical yield of 16.57 grams.
The actual yield from the experiment is 12.0 grams. To calculate the percent yield, we divide the actual yield by the theoretical yield and multiply by 100. This results in a percent yield of approximately 72.4 percent, indicating the efficiency of the reaction in a laboratory setting.
