Gọi \(\left\{{}\begin{matrix}n_{NaOH}=a\left(mol\right)\\n_{Ba\left(OH\right)_2}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)
\(\rightarrow n_{OH^-}=a+2b\left(mol\right)\)
\(m_{HCl}=\dfrac{350.7,3}{36,5}=25,55\left(g\right)\\ \rightarrow n_{H^+}=n_{HCl}=\dfrac{25,55}{36,5}=0,7\left(mol\right)\)
PTHH: \(OH^-+H^+\rightarrow H_2O\)
0,7<----0,7
\(\rightarrow a+2b=0,7\)
Theo bài ra, ta có hệ: \(\left\{{}\begin{matrix}40a+171b=50,75\\a+2b=0,7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,25\end{matrix}\right.\left(TM\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{NaOH}=\dfrac{0,2.40}{50,75}.100\%=15,76\%\\\%m_{Ba\left(OH\right)_2}=100\%-15,76\%=84,24\%\end{matrix}\right.\)
