Question

How many grams of barium chloride will be required to react with 5.0 gm of sodium sulfate

Answer 1

Answer:

The following chemical equation resulting in barium sulphate precipitate will be helpful in answering this question:

BaCl2 (soluble in waater) Any soluble sulphate or sulphuric acide will give white barium sulphate precipitate.

BaCl2 + Na2SO4 = BaSO4 + 2NaCl

In terms of weight (atomic):

Molecular wt. of BaCl2 = 137 +35.5 X 2 = 208, AND that of BaSO4 = 137 + 32 + 64

= 233.

Here 233 g of BaSO4 are obtained from 208 g of BaCl2, 5 g of BaSO4 will be obtained from:

(233 x 5) / 208 = 5.6 g of BaCl2

Explanation:

:D