a) \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
b)
\(n_{CuSO_4}=\dfrac{8}{160}=0,05\left(mol\right)\)
Theo PTHH: \(n_{Cu\left(OH\right)_2}=0,05\left(mol\right)\Rightarrow m_{Cu\left(OH\right)_2}=0,05.98=4,9\left(g\right)\)
c)
Theo PTHH: \(n_{NaOH}=0,1\left(mol\right)\Rightarrow m_{NaOH}=0,1.40=4\left(g\right)\)
=> \(m_{dd.NaOH}=\dfrac{4.100}{20}=20\left(g\right)\)