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Finding remaining mass of hydrate

Posted: Thu Oct 04, 2018 4:39 pm
by Emily Kennedy 4L
So if your given an Anhydrous copper (II) sulfate what amount would would remain after removing 90% of the water from 360 g of CuSO4.5H2O?

Re: Finding remaining mass of hydrate

Posted: Thu Oct 04, 2018 5:53 pm
by Jonathan Cheng 3C
(360 g CuSO4*5H2O) * (1 mol CuSO4*5H2O / 249.68 g CuSO4*5H2O) * (5 mol H2O / 1 mol CuSO4*5H2O) * (18 g H2O / 1 mol H2O) = 129.76 g H2O
This conversion gives the amount of water present in 360g of copper sulfate pentahydrate. When 90% of water is removed, then 116.79 g of H2O is removed.
360g - 116.79g = 243 g of remaining substance