a) Ta có: \(n_{NaOH}=\dfrac{2}{40}=0,05\left(mol\right)\)
PTHH: \(2NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)
0,05------>0,025----->0,025---------->0,05
=> mchất rắn = 0,025.58 = 1,45 (g)
b) \(\left\{{}\begin{matrix}C_{M\left(NaOH\right)}=\dfrac{0,025}{0,2}=0,125M\\C_{M\left(NaCl\right)}=\dfrac{0,05}{0,2}=0,25M\end{matrix}\right.\)