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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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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.

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