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.
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.
To determine if the K₂SO₄ will be sufficient to precipitate all the Ba⁺² ions in Ba(NO₃)₂, we first need to calculate the moles of Ba(NO₃)₂ and K₂SO₄ present in the reaction.
1. Start by calculating the moles of Ba(NO₃)₂:
- The concentration of Ba(NO₃)₂ is given as 0.10 M.
- The volume of Ba(NO₃)₂ used is 10.0 mL, which corresponds to 0.010 L.
- Multiply the concentration by the volume to get the moles:
Moles of Ba(NO₃)₂ = 0.10 M x 0.010 L = 0.0010 moles
2. Next, calculate the moles of K₂SO₄:
- The concentration of K₂SO₄ is given as 0.10 M.
- The volume of K₂SO₄ used is 20.0 mL, which corresponds to 0.020 L.
- Multiply the concentration by the volume to get the moles:
Moles of K₂SO₄ = 0.10 M x 0.020 L = 0.0020 moles
3. Determine the stoichiometric relationship between Ba(NO₃)₂ and K₂SO₄ in the balanced equation:
- The balanced equation shows that 1 mole of Ba(NO₃)₂ reacts with 1 mole of K₂SO₄ to produce 1 mole of BaSO₄ and 2 moles of KNO₃.
4. Now, compare the moles of Ba(NO₃)₂ and K₂SO₄:
- The mole ratio of Ba(NO₃)₂ to K₂SO₄ is 1:1. This means that for every 1 mole of Ba(NO₃)₂, we need 1 mole of K₂SO₄ to react completely.
- In our case, there are 0.0010 moles of Ba(NO₃)₂ and 0.0020 moles of K₂SO₄.
- Since the moles of K₂SO₄ are double the moles of Ba(NO₃)₂, there is an excess of K₂SO₄.
Therefore, the K₂SO₄ will be sufficient to precipitate all the Ba⁺² ions in the Ba(NO₃)₂ solution, and there will be excess K₂SO₄ remaining after the reaction is complete.