Question

750 mL of 0.250 M Na2SO4 solution are added to an aqueous solution containing 50 g of BaCl2 resulting in

the formation of white precipitate of BaSO4 according to following reaction. How many moles and grams of

barium sulphate will be obtained? ( Na= 23, S=32, O= 16, Ba=137, Cl=35.5)

BaCl2 + Na2SO4 → BaSO4 + 2Nacl.
tell the answer fast​

Answer 1

Answer:

0.24 moles or 56 gm of BaSo4