Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba (OH) 2 are mixed with 3.06 L of 0.0664 M Na2SO4.

We are given with two reactants : barium hydroxide and sodium sulfate. The products of the reaction via double replacement is barium sulfate and sodium hydroxide. according to the solubility rules the product barium sulfate is insoluble. There is a one to one correspondence to every compound. In this case, we just have to find the limiting reactant and base the calculations there. The mass of the precipitate formed is 0.186 grams BaSO4