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1)
Molar mass of CuSO₄•5H₂O = (63.5 + 32.1 + 16.0×9 + 1.0×10) g/mol = 249.6 g/mol
No. of moles of 30 g of CuSO₄•5H₂O = (30 g) / (249.6 g/mol) = 0.12 mol


2)
Percentage by mass of CuSO₄•5H₂O in the solution = [30/(30 + 150)] × 100(%) = 16.7%


3)
Molar mass of Na₂S = (23.0×2 + 32.1) g/mol = 78.1 g/mol
Initial number of moles of Na₂S = (8.8 g) / (78.1 g/mol) = 0.113 mol
Initial number of moles of CuSO₄ = Initial number of moles of CuSO₄•5H₂O = 0.12 mol

Solid Na₂S dissolves in water and then undergoes precipitate with CuSO₄.
Equation for the precipitation reaction :
CuSO₄(aq) + Na₂S(aq) → CuS(s) + Na₂SO₄(aq)
Mole ratio CuSO₄ : Na₂S : CuS = 1 : 1

When 0.113 mole of Na₂S completely reacts, CuSO₄ needed = 0.113 mol < 0.12 mol
Hence, 0.113 mole of Na₂S completely reacts, and CuSO₄ is in excess.

No. of moles of Na₂S reacted = 0.113 mol
No. of moles of CuS precipitate made = 0.113 mol

Molar mass of CuS = (63.5 + 32.1) g/mol = 95.6 g/mol
Mass of CuS precipitate made = (0.113mol) × (95.6 g/mol) = 10.8 g