Step-by-step explanation:
3 Ba(NO₃)₂ (aq) + Al₂(SO₄)₃ (aq) ----> 3 BaSO₄ (s) + 2 Al(NO₃)₃ (aq)
We have to find the volume in mL of a 0.380 M barium nitrate solution that we have to add to completely precipitate the sulfate ions from a 23.0 mL of a 0.250 M aluminum sulfate solution.
First we will have find the number of moles of aluminum sulfate that the 23.0 mL of 0.250 M solution contains.
Molarity = 0.250 M = 0.250 mol/L = 0.250 mmol/mL
mmoles of Al₂(SO₄)₃ = 23.0 mL * 0.250 mmol/mL
mmoles of Al₂(SO₄)₃ = 5.75 mmol
According to the coefficients of the reaction, 3 moles of Ba(NO₃)₂ will react with 1 mol of Al₂(SO₄)₃. Then the molar ratio between them is 3 to 1. We can use this relationship to find the number of mmoles of Ba(NO₃)₂ that will completely react with 5.75 mmol of Al₂(SO₄)₃.
3 mmol of Ba(NO₃)₂ = 1 mmol of Al₂(SO₄)₃
mmoles of Ba(NO₃)₂ = 5.75 mmol of Al₂(SO₄)₃ * 3 mmol of Ba(NO₃)₂/(1 mmol of Al₂(SO₄)₃)
mmoles of Ba(NO₃)₂ = 17.25 mmol
And finally we can find the volume of the 0.380 M solution of Ba(NO₃)₂ that we need.
volume of Ba(NO₃)₂ = 17.25 mmol/(0.380 mmol/mL)
volume of Ba(NO₃)₂ = 45.4 mL
Answer: 45.4 mL of the 0.380 M solution of Ba(NO₃)₂ are required.