Question

In an experiment 14.2 g of Sodium sulphate is allowed to react with 20.8 g of Barium chloride.
A white precipitate of Barium sulphate separates out which weighed 23.3 g and a solution of
Sodium chloride is formed. If the Law of conservation holds true for the chemical reaction,
calculate the mass of Sodium chloride formed

Answer 1

Answer:

Let the amount of barium sulphate is “x”g The given reaction is: Barium chloride + Sodium sulphate → Barium sulphate +sodium sulphate 20.8g 14.2g 11.7g According to law of conservation of mass, Amount of reactants = Amount of products formed Hence, 20.8g + 14.2g = x + 11.7g ⇒ 35g = x + 11.7g ⇒ x = 23.5g