Gọi \(\left\{{}\begin{matrix}n_{Ba}=a\left(mol\right)\\n_K=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{6,16}{22,4}=0,275\left(mol\right)\)
PTHH:
2K + 2H2O ---> 2KOH + H2
b b b/2
Ba + 2H2O ---> Ba(OH)2 + H2
a a a
Hệ pt \(\left\{{}\begin{matrix}137a+39b=33,25\\a+\dfrac{b}{2}=0,275\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,15\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}m_{Ba}=137.0,2=27,4\left(g\right)\\m_K=0,15.39=5,85\left(g\right)\end{matrix}\right.\\\left\{{}\begin{matrix}m_{Ba\left(OH\right)_2}=171.0,2=34,2\left(g\right)\\m_{KOH}=56.0,15=16,8\left(g\right)\end{matrix}\right.\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{27,4}{27,4+5,85}=82,4\%\\\%m_K=100\%-82,4\%=17,6\%\end{matrix}\right.\\\left\{{}\begin{matrix}\%m_{Ba\left(OH\right)_2}=\dfrac{34,2}{34,2+16,8}=67,05\%\\\%m_{KOH}=100\%-67,05\%=32,95\%\end{matrix}\right.\end{matrix}\right.\)
