A student adds 50.0 mL of 0.25 M K2SO4 solution to 25.0 mL 0.50 M BaCl2 solution. What is the mass of the precipitate that can be recovered if the percent yield of the reaction is 75%? Ba = 137;
Cl = 35.5; S = 32 0 = 16; K = 39. Note: You must write the balanced equation first.
BaCl2(aq) + K2SO4(aq) - BaSO4(s) + 2Cl(aq)