Asked by Janaiby

If a 8.6195 g of a mixture which contains (20.0 % of CuSO4.XH2O and 80.0 % of sand) was heated in
a crucible then cooled in the desiccator the mass obtained was 7.9976 g.
a- Find X the number of water of hydration in the Hydrate (CuSO4.XH2O)
(MM of CuSO4 = 159.5 g/mol and MM of H2O = 18.00 g/mol).
Answered by Janaiby
please help me ,becauce i have final exam tomorrow
Answered by DrBob222
8.6195 = g CuSO4.xH2O + sand
-7.9976 = g CuSO4 - H2O + sand
----------
0.6219 = g H2O lost

8.6195 = g CuSO4.xH2O + sand
x 0.20 CuSO.xH2O

= 1.7239 = g CuSO4.xH2O
-0.6219 = mass xH2O
--------------
= 1.102 = g CuSO4

mols CuSO4 = g/molar mass = 1.102/159.5 = estimated 0.0069
mols H2O = g/molar mass = 0.6219/18 = estimated 0.0345
Then find the ratio for mols H2O per 1 mol CuSO4. That's 0.0345/0.0069 = 5.0
Then find ratio of
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