- Ta có : \(m_{hh}=m_{Na}+m_{Ba}=7,09=23n_{Na}+137n_{Ba}\left(I\right)\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)
- Theo PTHH : \(n_{H_2}=\dfrac{V}{22,4}=0,075=\dfrac{1}{2}n_{Na}+n_{Ba}\left(II\right)\)
- Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na}=0,07\\n_{Ba}=0,04\end{matrix}\right.\) mol .\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,07\\n_{Ba\left(OH\right)_2}=0,04\end{matrix}\right.\) mol .
\(\Rightarrow n_{OH^-}=0,15mol\)
Theo bài ra : \(n_{H^+}=0,2V+2.0,15.V=0,5Vmol\)
PT : \(H^++OH^-\rightarrow H_2O\)
Theo PT ion : \(0,5V=0,15\)
\(\Rightarrow V=0,3\left(l\right)\)
- Ta lại có : \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)2}=0,04\\n_{H2SO4}=0,045\end{matrix}\right.\) mol
\(PTHH:Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)
Theo PTHH : \(m_{\downarrow}=m_{BaSO4}=0,04.M=9,32\left(g\right)\)
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