How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2NaCl (aq)

We are given the chemical reaction and the amount of the reactant to be used for the process. We use these data together with the molar masses to obtain what is asked. We do as as follows:

(0.113 mol / L) (.02534 L) (1 mol BaSO4 / 1 mol BaCl2) (233.38 g / mol) = 1.5273 g BaSO4 produced

Hope this answers the question.