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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?

PLEASE give the answer faster ​

Sagot :

pstnonsonjoku

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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