Question

Barium Sulfate precipitated from a 1.300 sample was contaminated with 9.4 mg of Fe2(SO4)3 and weighed a total of 314.0 mg. Calculate the apparent % BaSO4 and True % BaSO4
Apparent % BaSO4 = (314.0 - 9.4) / 1.300 x 100 = 24.15%

True % BaSO4 = (314.0 - 9.4) / (1.300 + 9.4) x 100 = 24.02%

Related Questions

An 800-mg sample containing sulfate was treated with slight excess of barium chloride, yielding a pr... 5.34g of salt of formula M2SO4 (M is a metal)were dissolved in water. The sulfate ion was precipitat... 150.0 mL of 0.105 M aluminium sulfate is mixed with 250.0 mL of 0.198 M barium chloride. What mass o... Barium sulfate (MW= 233.38 g/mole) precipitated from a 1.300 g sample was contaminated with 9.4 mg o...