How can i solve for this ? 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.

The balanced chemical reaction is:

Na2SO4 + Ba(OH)2 = 2NaOH +BaSO4

We are given the amount of the reactants to be used. These values will be the starting point of our calculations.

0.0820 mol/L Ba(OH)2 (2.27 L solution) = 0.18614 mol Ba(OH)2
0.0664 mol/L Na2SO4 (3.06 L solution) = 0.20318 mol Na2SO4

The limiting reactant is the Ba(OH)2. The amount for this compound will be used.

0.18614 mol Ba(OH)2 (1 mol BaSO4 / 1 mol Ba(OH)2 ) (233.43 g BaSO4 / 1 mol BaSO4)= 43.4507 g BaSO4