Question #87321

Assuming that the sodium sulphate sample is 100% pure, calculate the volume of 0.25 M BaCL2 required to precipitate all the sulphate in it. (as in step 4 in the precipitation stage in the experiment). What is the formula use for this question ? thank you.

Expert's answer

1 mol X mol

Na_{2}SO_{4} +BaCl_{2} = BaSO_{4}↓ + 2NaCl

1 mol 1 mol

n(Na_{2}SO_{4}) = 1 mol = 142 g/mol.

n(BaCl_{2}) = X = 1 mol.

V(BaCl_{2}) = n(BaCl_{2}) / C(BaCl_{2}) = 1 mol / 0.25 M = 4 L.

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