\(NaOH+HCl\rightarrow NaCl+H_2O\left(1\right)\\ KOH+HCl\rightarrow KCl+H_2O\left(2\right)\\ Tathấyởpứ\left(1\right),\left(2\right):n_{H_2O}=n_{HCl}=0,2\left(mol\right)\\ BTKL:m_{hh.hidroxit}+m_{HCl}=m_{muối}+m_{H_2O}\\ \Rightarrow m_{hh.hidroxit}=13,3+0,2.18-0,2.36,5=9,6\left(g\right)\\ Đặt:\left\{{}\begin{matrix}n_{NaOH}=x\left(mol\right)\\n_{KOH}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}40x+56y=9,6\\58,5x+74,5=13,3\end{matrix}\right.\\\Rightarrow \left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\\ \Rightarrow\%NaOH=41,67\%;\%KOH=58,33\%\)
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