Can someone help me with chemistry, solubility questions? Thank u?

6.
A)
CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)


B)
Molar mass of CuSO₄ = (63.5 + 32.1 + 16.0×4) g/mol = 159.6 g/mol
Initial no. of moles of CuSO₄ = [(5.80 g) / (159.6 g/mol)] × (25.0/125) = 0.007268 mol
Initial no. of moles of NaOH = (0.125 mol/L) × (125/1000 L) = 0.01563 mol

According to the above equation, mole ratio CuSO₄ : NaOH = 1 : 2
If all the 0.007268 mol CuSO₄ completely reacts, NaOH needed = (0.007268 mol) × 2 = 0.01454 mol < 0.01563 mol
Hence, NaOH is in excess.
In other words, CuSO₄ is the limiting reactant.


C)
According to the above equation, mole ratio CuSO₄ : Cu(OH)₂ = 1 : 1
No. of moles of CuSO₄ reacted = 0.007268 mol
No. of moles of Cu(OH)₂ precipitated = 0.007268 mol

Molar mass of Cu(OH)₂ = (63.5 + 16.0×2 + 1.0×2) g/mol = 97.5 g/mol
Mass of Cu(OH)₂ precipitated = (0.007268 mol) × (97.5 mol) = 0.709 g