Calculate the mass of barium sulfate (BaSO4, molar mass = 233.43 g/mol) that can be produced when 350.0 mL of a 0.100-M solution of barium chloride (BaCl2) is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate (Fe2(SO4)3)? a. 2.33 g b. 7.00 g c. 3.50 g d. 2.85 g
The Correct Answer and Explanation is:
Correct Answer: a. 2.33 g
Step-by-step Calculation:
We begin with the balanced chemical reaction:
3BaCl₂ (aq) + Fe₂(SO₄)₃ (aq) → 3BaSO₄ (s) + 2FeCl₃ (aq)
From the equation, we see that 3 moles of BaCl₂ react with 1 mole of Fe₂(SO₄)₃ to form 3 moles of BaSO₄.
Step 1: Calculate moles of each reactant
BaCl₂ solution:
Volume = 350.0 mL = 0.350 L
Molarity = 0.100 mol/L
Moles of BaCl₂ = 0.100 mol/L × 0.350 L = 0.0350 mol
Fe₂(SO₄)₃ solution:
Volume = 100.0 mL = 0.100 L
Molarity = 0.100 mol/L
Moles of Fe₂(SO₄)₃ = 0.100 mol/L × 0.100 L = 0.0100 mol
Step 2: Determine limiting reactant
From the balanced equation:
- 3 mol BaCl₂ reacts with 1 mol Fe₂(SO₄)₃
- So, 0.0350 mol BaCl₂ would require:
0.0350 mol ÷ 3 = 0.0117 mol Fe₂(SO₄)₃
But we only have 0.0100 mol Fe₂(SO₄)₃, which is less than 0.0117 mol.
This means Fe₂(SO₄)₃ is the limiting reactant.
Step 3: Calculate moles of BaSO₄ formed
From the balanced equation:
1 mol Fe₂(SO₄)₃ forms 3 mol BaSO₄
Therefore:
0.0100 mol Fe₂(SO₄)₃ × 3 = 0.0300 mol BaSO₄
Step 4: Convert moles of BaSO₄ to grams
Molar mass of BaSO₄ = 233.43 g/mol
Mass = 0.0300 mol × 233.43 g/mol = 7.0029 g
Wait — that’s not right. We must have made an error in assigning the limiting reactant.
Let’s go back:
Correction:
If BaCl₂ is 0.0350 mol, and the reaction requires 3 mol BaCl₂ per 1 mol Fe₂(SO₄)₃, then the Fe₂(SO₄)₃ needed = 0.0350 ÷ 3 = 0.0117 mol
But we only have 0.0100 mol of Fe₂(SO₄)₃ → so Fe₂(SO₄)₃ is still limiting
So the moles of BaSO₄ = 0.0100 mol Fe₂(SO₄)₃ × 3 = 0.0300 mol
And the mass = 0.0300 mol × 233.43 g/mol = 7.0029 g
Therefore, correct answer is b. 7.00 g
The initially marked answer a. 2.33 g was incorrect — apologies for the confusion.
Final Answer: b. 7.00 g
This problem emphasizes the importance of understanding stoichiometry, limiting reactants, and molar mass relationships in precipitation reactions. When two solutions react to form a precipitate such as BaSO₄, the amount formed is governed by the amount of limiting reactant present. In this case, the iron(III) sulfate limits the formation of barium sulfate. Using the balanced equation and stoichiometric ratios, we determined the amount of BaSO₄ produced in grams accurately.
