PTHH: \(2NaOH+MgCl_2\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=\frac{40\cdot20\%}{40}=0,2\left(mol\right)\\n_{MgCl_2}=\frac{190\cdot10\%}{95}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\frac{0,2}{2}< \frac{0,2}{1}\) \(\Rightarrow\) NaOH phản ứng hết, MgCl2 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{Mg\left(OH\right)_2}=0,1mol=n_{MgCl_2\left(dư\right)}\\n_{NaCl}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Mg\left(OH\right)_2}=0,1\cdot58=5,8\left(g\right)\\m_{MgCl_2\left(dư\right)}=0,1\cdot95=9,5\left(g\right)\\m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddNaOH}+m_{ddMgCl_2}-m_{Mg\left(OH\right)_2}=224,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2\left(dư\right)}=\frac{9,5}{224,2}\cdot100\approx4,24\%\\C\%_{NaCl}=\frac{11,7}{224,2}\cdot100\approx5,22\%\end{matrix}\right.\)