\(n_{Na}=\frac{4,6}{23}=0,2\left(mol\right)\)
\(n_{Ca}=\frac{10}{40}=0,25\left(mol\right)\)
PTHH: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
_______0,2-------------------->0,2------>0,1___(mol)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
0,25----------------->0,25----->0,25____(mol)
=> \(\left\{{}\begin{matrix}m_{NaOH}=0,2.40=8\left(g\right)\\m_{Ca\left(OH\right)_2}=0,25.74=18,5\left(g\right)\\m_{dd}=4,6+10+150-2\left(0,1+0,25\right)=163,9\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C\%\left(NaOH\right)=\frac{8}{163,9}.100\%=4,88\%\\C\%\left(Ca\left(OH\right)_2\right)=\frac{18,5}{163,9}.100\%=11,29\%\end{matrix}\right.\)
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