[tarik3001]

The determination of weight % SO42- by precipitation with Ba2+ ion to form the insoluble BaSO4 is a well-known method in the wet chemical analysis laboratory.  The reaction proceeds as described by the equation

Ba2+ (aq) +  SO42-  (aq)     to     BaSO4 (s)

A 0.5217 g sample of pure K2SO4 is treated with 30.00 mL of 0.1508 M BaCl2, with the formation of the BaSO4 precipitate.  Determine if enough BaCl2 was added to the solution to precipitate all of the sulfate from the pure potassium sulfate sample.

did I do it right..

.030000L * .1508M BaCl2 = .004524M BaCl2
.5217g K2SO4 * (1/174.262) = .002994 M K2SO4 required

[opti384]

It seems that you got the calculations right but be careful with "M" because it usually means molarity.

[Borek]

Just remember you have not answered the question yet.

[Cherriyan]

K₂SO₄ + BaCl₂  BaSO₄ + 2KCl

Since you need to determine if the volume of BaCl₂ added is enough for precipitation of sulphate, you should calculate the number of moles of K₂SO₄ present and BaCl₂ added, then compare the two. If number of moles of BaCl₂ >= those of K₂SO₄, then the BaCl₂ is sufficient to carry out the reaction.
According to my calculations, no. of moles of K₂SO₄ = 0.002998
And no. of moles of BaCl₂ is coming ouit to be 0.004524, which is definitely greater than the value mentioned above. So BaCl₂ is sufficient to carry out the above precipitation reaction (i.e., sulphate has been completely precipitated)