Question

A student was given a 2.45 g of impure potassium sulfate (K2SO4) sample. He dissolved this sample in 100 mL of water first and then added some barium chloride (BaCl2) solution. A 2.15 g of BaSO4 precipitate was obtained at the end of the reaction. What is the percent purity of K2SO4 in the original sample? (

Answer 1

mols BaSO4 = grams/molar mass
Convert mols BaSO4 to mols K2SO4. The easy way to do this is to recognize there is 1 mol SO4 in BaSO4 and 1 mol in K2SO4 so mols BaSO4 = mols K2SO4.
Then convert mols K2SO4 to grams. g = mols x molar mass
Next convert mols K2SO4 to grams K2SO4. g = mols x molar mass
Finally, %K2SO4 = % purity = (g K2SO4/mass sample)*100 = ?

Answer 2

thank u