Stoichiometry and Limiting Reagent in a Precipitation Reaction
A technician mixes 250.0 mL of 0.200 mol/L barium chloride solution with 150.0 mL of 0.300 mol/L sodium sulfate solution. A white precipitate forms immediately. The technician filters, dries and weighs the precipitate, obtaining 6.50 g. She is asked to check whether this mass is reasonable, to identify the limiting reagent, and to state the concentration of the spectator ions remaining in the filtrate (assume the precipitate is completely removed and volumes are additive). M(BaSO4) = 233 g/mol.
1. Write the net ionic equation, identify the limiting reagent, and calculate the theoretical mass of precipitate.
Read the worked answer
Net ionic equation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
Moles of Ba²⁺ = 0.2500 L × 0.200 mol/L = 0.0500 mol. Moles of SO₄²⁻ = 0.1500 L × 0.300 mol/L = 0.0450 mol.
The reaction is 1:1, so the reagent with fewer moles is limiting: sulfate (SO₄²⁻) is limiting, Ba²⁺ is in excess by 0.0500 − 0.0450 = 0.0050 mol.
Theoretical moles of BaSO₄ = 0.0450 mol. Theoretical mass = 0.0450 × 233 = 10.5 g.
The measured 6.50 g is only about 62% of the theoretical yield. That is far too low for a clean precipitation. Likely causes: incomplete precipitation, loss during filtration, the precipitate was not fully dried, or the solution was not at the assumed concentrations. The mass is not reasonable as a quantitative yield and the procedure should be checked.
2. Calculate the concentrations of the spectator ions (Na⁺ and Cl⁻) in the final filtrate, given total volume is 400.0 mL.
Read the worked answer
Spectator ions are unchanged: Na⁺ from sodium sulfate and Cl⁻ from barium chloride.
Moles Na⁺ = 2 × moles Na₂SO₄ = 2 × 0.0450 = 0.0900 mol. Moles Cl⁻ = 2 × moles BaCl₂ = 2 × 0.0500 = 0.100 mol.
Total volume = 250.0 + 150.0 = 400.0 mL = 0.4000 L.
[Na⁺] = 0.0900 / 0.4000 = 0.225 mol/L. [Cl⁻] = 0.100 / 0.4000 = 0.250 mol/L.
Note that Ba²⁺ is not a spectator in the same sense - the excess 0.0050 mol remains in solution: [Ba²⁺] = 0.0050 / 0.4000 = 0.0125 mol/L (assuming all sulfate precipitated). A careful answer should mention this residual Ba²⁺, since it affects charge balance: 2[Ba²⁺] + [Na⁺] should equal [Cl⁻] plus any residual sulfate. Checking: 2(0.0125) + 0.225 = 0.250 = [Cl⁻], consistent.