A solution contains 0.0100 mol $Na X_{2} SO X_{4}$ and 0.0100 mol $Na X_{3} PO X_{4}$ . Excess $BaCl X_{2}$ is added at neutral pH, forming both barium salts completely, and then excess dilute $HCl$ is added. The mass of the precipitate remaining is (Ba 137, S 32, P 31, O 16)

Answer: A

BaSO X_{4}

(

233 g mo l^{- 1}

) is 0.0100 mol, i.e. 2.33 g.

Ba X_{3} (PO X_{4}) X_{2}

(601) would be 0.00500 mol, 3.01 g, so the initial precipitate is 5.34 g. In excess dilute

HCl

the phosphate dissolves and only

BaSO X_{4}

, 2.33 g, remains.