Question

mixture of CuSO4 5H2O and MgSO4 7H2O is heated until all the water is lost.

Answer 1

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so here's your answer!!☺☺
CuSO4*5H2O -----> CuSO4 + 5H2O
MgSO4* 7H2O -----> MgSO4 + 7H2O
mass of water = 5.020g - 2.988g = 2.032g
moles of H2O = 2.032g(1mole/18g) = 0.113 mole
in the mixture there are 12 moles of H2O

moles of H2O from CuSO4*5H2O
(5/12)*0.113 = 0.047 mole
moles of H2O from MgSO4*5H2O
(7/12)*0.113 = 0.066 mole

since there is 1:5 mole ratio between CuSO4*5H2O&H2O then mole of CuSO4 = 0.047/5 = 0.0093 mole CuSO4*5H2O
the MM of CuSO4*5H2O is 249.69 g/mol
the mass of CuSO4*5H2O then is
0.0093mole*249.69 g/mol = 2.322 g
the percent by mass of the CuSO4*5H2O is then
(2.322g/5.020g)*100 = 46.25%

hope it helps you!!☺☺

Answer 2

Answer:

74%

Explanation:

By applying POAC we equate the moles of water on the reactant side and product side.