a) \(n_{MgCl_2}=\dfrac{10}{95}=\dfrac{2}{19}\left(mol\right)\)
PTHH: \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)
\(\dfrac{2}{19}\)----->\(\dfrac{4}{19}\)---------->\(\dfrac{2}{19}\)
b) => \(m_{kt}=\dfrac{4}{19}.58=\dfrac{232}{19}\left(g\right)\)
c) \(C_{M\left(NaOH\right)}=\dfrac{\dfrac{2}{19}}{0,2}=\dfrac{10}{19}M\)