If 8.753 g of the heptahydrate produces 8.192 g of the hexahydrate, how many grams of anhydrous nickel (ii) sulfate could be obtained?

Question

Haylee Ford

Chemistry

28 August, 19:43

If 8.753 g of the heptahydrate produces 8.192 g of the hexahydrate, how many grams of anhydrous nickel (ii) sulfate could be obtained?

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Dominguez · Answer 1

Answer is: 4.826 grams of anhydrous nickel (II) sulfate could be obtained.

m (NiSO₄*7H₂O) = 8.753 g; mass of heptahydrate.

m (NiSO₄*6H₂O) = 8.192 g; mass of hexahydrate.

m (H₂O) = m (NiSO₄*7H₂O) - m (NiSO₄*6H₂O).

m (H₂O) = 8.753 g - 8.192 g.

m (H₂O) = 0.561 g.

m (NiSO₄) = m (NiSO₄*7H₂O) - 7 · m (H₂O).

m (NiSO₄) = 8.753 g - 7 · 0.561 g.

m (NiSO₄) = 4.826 g.