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 :
To determine the mass of barium sulfate (BaSO4\text{BaSO}_4) produced, we must follow a step-by-step stoichiometric analysis of the reaction:
Step 1: Write the Balanced Chemical Equation
The reaction between barium chloride (BaCl2\text{BaCl}_2) and iron(III) sulfate (Fe2(SO4)3\text{Fe}_2(\text{SO}_4)_3) is: 3BaCl2+Fe2(SO4)3→3BaSO4↓+2FeCl33\text{BaCl}_2 + \text{Fe}_2(\text{SO}_4)_3 \rightarrow 3\text{BaSO}_4 \downarrow + 2\text{FeCl}_3
This equation shows that 3 moles of BaCl₂ react with 1 mole of Fe₂(SO₄)₃ to form 3 moles of BaSO₄.
Step 2: Determine the Moles of Reactants
Barium chloride (BaCl2\text{BaCl}_2)
- Molarity (M) = 0.100 M
- Volume = 350.0 mL = 0.350 L
- Moles of BaCl₂: 0.100 M×0.350 L=0.0350 moles of BaCl20.100 \text{ M} \times 0.350 \text{ L} = 0.0350 \text{ moles of BaCl}_2
Iron(III) sulfate (Fe2(SO4)3\text{Fe}_2(\text{SO}_4)_3)
- Molarity (M) = 0.100 M
- Volume = 100.0 mL = 0.100 L
- Moles of Fe₂(SO₄)₃: 0.100 M×0.100 L=0.0100 moles of Fe2(SO4)30.100 \text{ M} \times 0.100 \text{ L} = 0.0100 \text{ moles of Fe}_2(\text{SO}_4)_3
Step 3: Determine the Limiting Reactant
From the balanced equation:
- 3 moles of BaCl₂ react with 1 mole of Fe₂(SO₄)₃.
- The available 0.0350 moles of BaCl₂ would need: 0.0350 moles BaCl23=0.0117 moles Fe2(SO4)3\frac{0.0350 \text{ moles BaCl}_2}{3} = 0.0117 \text{ moles Fe}_2(\text{SO}_4)_3
- However, only 0.0100 moles Fe₂(SO₄)₃ are available, which is less than required.
Thus, Fe₂(SO₄)₃ is the limiting reactant.
Step 4: Calculate the Moles of BaSO₄ Formed
From the reaction, 1 mole of Fe₂(SO₄)₃ produces 3 moles of BaSO₄: 0.0100 moles Fe2(SO4)3×3 moles BaSO41 mole Fe2(SO4)3=0.0300 moles BaSO40.0100 \text{ moles Fe}_2(\text{SO}_4)_3 \times \frac{3 \text{ moles BaSO}_4}{1 \text{ mole Fe}_2(\text{SO}_4)_3} = 0.0300 \text{ moles BaSO}_4
Step 5: Calculate the Mass of BaSO₄
- Molar mass of BaSO₄ = 233.43 g/mol
- Mass: 0.0300 moles×233.43 g/mol=7.00 g0.0300 \text{ moles} \times 233.43 \text{ g/mol} = 7.00 \text{ g}
Final Answer:
7.00 g(Option b)\mathbf{7.00 \text{ g} \quad (Option \; b)}