The equation shows that 1 mole of K₂SO₄ is needed to react with mole of Ba ions.
Just eyeballing the numbers that you provided, the K₂SO₄ that you provided is in excess compared to Ba ions in the reaction. Therefore, the amount of K₂SO₄ will be sufficient to precipitate Ba.
If you need to know how to do this calculation wise, Molarity*volume in L=moles
convert mL to L by multiplying the volume by 10^-3.
moles of K₂SO₄>Ba (NO₃)₂, Ba will precipitate.
Consider
Ba (NO₃)₂ + K₂SO₄ → BaSO₄ + 2KNO₃
(0.10 M) (0.10 M)
(10.0 mL) (20.0 mL)
Will the K₂SO₄ be sufficient to precipitate all the Ba⁺² ions in the Ba (NO₃)₂? Use Stoichiometry to answer this question.
1 answer
1 answer