Câu 3
\(n_M=\dfrac{26.91}{M_M}\left(mol\right)\)
\(n_{AlCl3}=0,5.0,7=0.35\left(mol\right)\)
\(n_{Al\left(OH\right)3}=\dfrac{17,94}{78}=0,23\left(mol\right)\)
Do \(n_{AlCl3}>n_{Al\left(OH\right)3}\rightarrow\)Kết tủa Al(OH)3 bị hòa tan
Ta có các PTHH
\(2M+2H2O\rightarrow2MOH+H2\left(1\right)\)
\(AlCl3+3MOH\rightarrow Al\left(OH\right)3+3MCl\left(2\right)\)
\(Al\left(OH\right)3+MOH\rightarrow MAlO2+2H2O\left(3\right)\)
Xác định kim loại M
\(\left(1\right)\rightarrow n_{MOH\left(bandau\right)}=n_M=\dfrac{26,91}{M_M}\left(mol\right)\)
\(\left(2\right)\rightarrow\left\{{}\begin{matrix}n_{Al\left(OH\right)3}=n_{AlCl3}=0.35\left(mol\right)\\\\n_{MOH\left(tạo ra \downarrow\right)}=n_{AlCl3}.3=1.05\left(mol\right)\end{matrix}\right.\)
\(\rightarrow n_{MOH\left(hòatan\downarrow\right)}=\dfrac{26,91}{M_M}-1.05\left(mol\right)\)
\(\left(3\right)\rightarrow n_{Al\left(OH\right)3\left(bi hòa tan\right)}=n_{MOH\left(hòa tan \downarrow\right)}=\dfrac{26,91}{M_M}-1.05\left(mol\right)\)
Do số mol kết tủa sau cùng thu được là 0,23mol , ta có pt:
\(0.35-\left(\dfrac{26,91}{M_M}-1.05\right)=0.23\Leftrightarrow M_M=23\left(\dfrac{g}{mol}\right)\rightarrow M:Na\)
Tính V
\(\left(1\right)\rightarrow n_{H2}=\dfrac{n_M}{2}=\dfrac{\dfrac{26,91}{23}}{2}=0.585\left(mol\right)\rightarrow V_{H2}=13,104\left(mol\right)\)