Question

find the mass of barium sulfate formed when 30. mL of 0.2 M barium nitrate is mixed with 100 mL of 0.04 M sodium sulfate

Answer 1

Na2SO4 + Ba(NO3)2 ----> 2 NaNo3 + BaSO4

moles of sodium sulfate = 0.04*0.1 = 4*10^-3 moles

moles of Barium nitrate = .03*0.2 = 6*10^-3 moles

so the limiting reagent is Na2SO4

so moles of BaSO4 formed = 4 millimoles

mass = 4*10^-3 * 233 = 0.932 grams