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. ​

Answer 1

Explanation:

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Answer 2

Answer:

Explanation:

moles of na2so4 = 0.25*0.75=0.1875

moles of bacl2=0.2404

limiting ratio = 0.1875

moles of baso4 = 0.1875

mass = 43.6875 gm