A mixture of unknown salts containing a sulfate salt has been weighed to be 1.50 g. After dissolving the solution in 500 mL of water, heating it, and adding hot barium chloride (40.0 mL, 0.500 M); the precipitate has been collected after succeeding digestion, and the mass was found to be 0.325 g. What is the percentage (by mass) of sulfur in the mixture?
BaCl2 + SO42- = BaSO4 + 2Cl-
The precipitate is BaSO4;
M(BaSO4) = 233 g/mole;
n(BaSO4) = m/M = 0.325/233 = 0.0014 moles;
According to the chemical reaction n(BaCl2) = n(SO42-) = n(BaSO4) = 0.0014 moles.
According to the task n(BaCl2) = C(BaCl2) * V(BaCl2) = 0.5 * 0.04 = 0.02 moles.
Therefore, salt BaCl2 is in excess.
n(SO42-) = n(S) = 0.0014 moles;
m(SO42-) = n(SO42-) * M(SO42-) = 0.0014 * 96 = 0.1344 g;
m(S) = n(S) * M(S) = 0.0014 * 32 = 0.0448 g;
X(SO42-) = m(SO42-) * 100%/m(mixture) = 0.1344 * 100%/1.5 = 9.0%;
X(S) = m(S) * 100%/m(mixture) = 0.0448 * 100%/1.5 = 3.0%.
Answer: 3.0%
Comments
Leave a comment