How many grams of water are evaporated by heating 50g of copper II sulfate penta hydrate ?

Mass of H₂O in 1 mole of CuSO₄•5H₂O = 5 × (1×2 + 16) g = 90 g
Mass of 1 mole of CuSO₄•5H₂O = (64 + 32 + 16×4 + 90) g = 250 g
Mass fraction of H₂O in CuSO₄•5H₂O = 90/250

Mass of H₂O evaporated by heating 50 g of CuSO₄•5H₂O = (50 g) × (90/250) = 18 g

50 g CuSO4-5H20 x 1 mole/250 g = 0.2 moles
0.2 moles CuSO4-5H2O x 5 moles H2O/mole CuSO4-5H2O = 1 mole H2O
1 mole H2O x 18 g/mole = 18 g H2O evaportaed

Molar mass of CuSO4*5H2O = 249.6850 g/mol
Mol CuSO4 in 50g = 50/149.6850 = 0.20 mol
1mol CuSO4.5H2O contains 5 mol H2O
0.20 mol CuSO4.5H2O contains 0.20*5 = 1.0 mol H2O
Molar mass H2O = 18g/mol
Mass of 1.0 mol H2O = 1.0*18 = 18g H2O evaporated.