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Hello,

Very simply, if the molar ratio is 1:1 mole, then divide the molecular weight of the anhydrous material by the hydrated one, then multiply the result by 8.753 g.

Solution:

NiSO4 anhydrous molecular weight is 154.75 g/mol (anhydrous)

NiSO4·7H2O molecular weight is 280.86 g/mol (heptahydrate)

So: 154.75/280.86 = 0.55098

The grams quantity produced from 8.753 heptahydrate is (0.55098*8.753) = 4.823 grams anhydrous NiSO4 anhydrous

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Q: How many grams of anhydrous nickel 2 sulfate is produced from 8.753 grams of heptahydrate?
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