Ba(NO3)2 + K2SO4 --> 6BaSO4 + 2KNO3 ; Will the k2SO4 be sufficient to precipitate all the Ba2+ ions in the Ba(NO3)2? Use stoichiometry to answer this question.

Answer 1

The reaction you provided as written is incorrect. You do NOT produce 6BaSO4. I have written the correctly balance equation below:

Correctly balanced equation:

Ba(NO₃)₂(aq) + K₂SO₄(aq) --> BaSO₄(s) + 2KNO₃(aq) ... balanced equation

There is not enough information provided to determine if the K₂SO₄ is sufficient to precipitate all of the Ba²⁺ ions because we don't know how much Ba²⁺ is present and we don't know how much K₂SO₄ is present. There must be more information that you've omitted.

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Edited response assuming the following:

10.0 ml of 0.10 M Ba(NO₃)₂ and 20.0 ml of 0.10 M K₂SO₄

Ba(NO₃)₂(aq) + K₂SO₄(aq) --> BaSO₄(s) + 2KNO₃(aq) ... balanced equation

moles Ba(NO₃)₂ present = 10.0 ml x 1 L / 1000 ml x 0.10 mol/L = 0.001 mols Ba(NO₃)₂ present

moles K₂SO₄ present = 20.0 ml x 1 1 L / 1000 ml x 0.10 mol/L = 0.002 moles K₂SO₄ present

Since the stoichiometry of the balanced equation shows it takes equal number of moles of each reactant (i.e 1:1 mol ratio), there is more than enough (twice as much) K₂SO₄ to precipitate all of the Ba²⁺ in Ba(NO₃)₂.