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What weight of barium chloride will react with 2.36g of sodium sulphate in solution so as to produce 3.88g of barium sulphate and 1.94g of sodium chloride in solution?

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Answer:

3.46 g of BaCl2

Explanation:

The equation of the reaction is;

BaCl2(aq) + Na2SO4(aq) --------> 2NaCl(aq) + BaSO4(s)

Number of moles of barium sulphate produced = 3.88 g/233.38 g/mol = 0.0166 moles

From the reaction equation;

1 mole of BaCl2 yields 1 mole of barium sulphate

Hence 0.0166 moles of BaCl2 yields 0.0166 moles of barium sulphate

Hence;

Mass of BaCl2 required = 0.0166 moles × 208.23 g/mol = 3.46 g of BaCl2

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