a) PTHH: \(CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
b) Ta có: \(n_{NaOH}=\frac{100\cdot10\%}{40}=0,25\left(mol\right)\)
Xét tỉ lệ: \(\frac{0,1}{1}< \frac{0,25}{2}\) \(\Rightarrow\) CuSO4 phản ứng hết, NaOH còn dư
\(\Rightarrow n_{Cu\left(OH\right)_2}=0,1mol\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,1\cdot98=9,8\left(g\right)\)
c) Theo PTHH: \(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,1mol\\n_{NaOH\left(dư\right)}=0,05mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Na_2SO_4}=0,1\cdot142=14,2\left(g\right)\\m_{NaOH\left(dư\right)}=0,05\cdot40=2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddCuSO_4}+m_{ddNaOH}-m_{Cu\left(OH\right)_2}=200\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2SO_4}=\frac{14,2}{200}\cdot100=7,1\%\\C\%_{NaOH\left(dư\right)}=\frac{2}{200}\cdot100=1\%\end{matrix}\right.\)