What mass of barium sulfate (molar mass = 233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of a 0.150 M solution of iron(III) sulfate?
The correct answer and explanation is:
The chemical reaction between barium chloride (BaCl₂) and iron(III) sulfate (Fe₂(SO₄)₃) forms barium sulfate (BaSO₄) and iron(III) chloride (FeCl₃) as products. The balanced equation for this reaction is: 3 BaCl2(aq)+Fe2(SO4)3(aq)→3 BaSO4(s)+2 FeCl3(aq)3 \, \text{BaCl}_2 (aq) + Fe_2(SO_4)_3 (aq) \rightarrow 3 \, \text{BaSO}_4 (s) + 2 \, \text{FeCl}_3 (aq)
Step 1: Moles of barium chloride and iron(III) sulfate
We are given:
- Volume of barium chloride solution = 125 mL = 0.125 L
- Concentration of barium chloride solution = 0.150 M
- Volume of iron(III) sulfate solution = 125 mL = 0.125 L
- Concentration of iron(III) sulfate solution = 0.150 M
To find the moles of each reactant, we use the formula: Moles=Molarity×Volume (L)\text{Moles} = \text{Molarity} \times \text{Volume (L)}
For barium chloride: Moles of BaCl2=0.150 M×0.125 L=0.01875 moles\text{Moles of BaCl}_2 = 0.150 \, \text{M} \times 0.125 \, \text{L} = 0.01875 \, \text{moles}
For iron(III) sulfate: Moles of Fe2(SO4)3=0.150 M×0.125 L=0.01875 moles\text{Moles of Fe}_2(SO_4)_3 = 0.150 \, \text{M} \times 0.125 \, \text{L} = 0.01875 \, \text{moles}
Step 2: Limiting reactant
From the balanced equation, the stoichiometric ratio between barium chloride and iron(III) sulfate is 3:1. For every 3 moles of BaCl₂, 1 mole of Fe₂(SO₄)₃ is required. Since both reactants are present in equal amounts (0.01875 moles), iron(III) sulfate will be the limiting reactant because only 1 mole of Fe₂(SO₄)₃ is required for every 3 moles of BaCl₂.
Step 3: Moles of barium sulfate produced
Since the limiting reactant is iron(III) sulfate, the number of moles of barium sulfate produced will be determined by the moles of Fe₂(SO₄)₃. From the balanced equation, 1 mole of Fe₂(SO₄)₃ produces 3 moles of BaSO₄. Moles of BaSO4=3×Moles of Fe2(SO4)3=3×0.01875=0.05625 moles\text{Moles of BaSO}_4 = 3 \times \text{Moles of Fe}_2(SO_4)_3 = 3 \times 0.01875 = 0.05625 \, \text{moles}
Step 4: Mass of barium sulfate
To calculate the mass of barium sulfate, we multiply the number of moles of BaSO₄ by its molar mass: Mass of BaSO4=Moles of BaSO4×Molar Mass of BaSO4\text{Mass of BaSO}_4 = \text{Moles of BaSO}_4 \times \text{Molar Mass of BaSO}_4 Mass of BaSO4=0.05625 moles×233 g/mol=13.12 g\text{Mass of BaSO}_4 = 0.05625 \, \text{moles} \times 233 \, \text{g/mol} = 13.12 \, \text{g}
Final Answer:
The mass of barium sulfate produced is 13.12 g.