50.0 mL of 0.200 M BaCl2 is mixed with 100.0 mL of 0.150 M Na2SO4. Assuming complete precipitation of BaSO4 and additive volumes, the mass of precipitate and the final [SO4^2-] are (Ba = 137, S = 32, O = 16)

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JEE Advanced levelSingle correct optionDifficulty 3 of 3mc196

$50.0 mL$ of $0.200 M$ $BaCl X_{2}$ is mixed with $100.0 mL$ of $0.150 M$ $Na X_{2} SO X_{4}$ . Assuming complete precipitation of $BaSO X_{4}$ and additive volumes, the mass of precipitate and the final $[SO X_{4} X^{2 -}]$ are (Ba = 137, S = 32, O = 16)

A $3.49 g$ and $0 M$
B $2.33 g$ and $0.050 M$
C $2.33 g$ and $0.0333 M$
D $2.33 g$ and $0.100 M$

Answer: C

n(Ba2+) = 0.0500 x 0.200 = 0.0100 mol; n(SO4 2-) = 0.100 x 0.150 = 0.0150 mol, so Ba2+ is limiting. BaSO4 = 0.0100 x 233 = 2.33 g. Sulfate left = 0.0050 mol in 0.150 L, i.e. 0.0333 M.

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50.0 mL of 0.200 M BaClX2​ is mixed with 100.0 mL of 0.150 M NaX2​SOX4​. Assuming complete precipitation of BaSOX4​ and additive volumes, the mass of precipitate and the final [SOX4​X2−] are (Ba = 137, S = 32, O = 16)

Answer: C

More questions on Limiting reagent, percent yield and percent excess

$50.0 mL$ of $0.200 M$ $BaCl X_{2}$ is mixed with $100.0 mL$ of $0.150 M$ $Na X_{2} SO X_{4}$ . Assuming complete precipitation of $BaSO X_{4}$ and additive volumes, the mass of precipitate and the final $[SO X_{4} X^{2 -}]$ are (Ba = 137, S = 32, O = 16)