PTHH: \(R+2H_2O\rightarrow R\left(OH\right)_2+H_2\uparrow\)
a) Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)=n_R\) \(\Rightarrow M_R=\dfrac{20,55}{0,15}=137\)
\(\Rightarrow\) Kim loại cần tìm là Bari
c) PTHH: \(3Ba\left(OH\right)_2+Al_2\left(SO_4\right)_3\rightarrow2Al\left(OH\right)_3\downarrow+3BaSO_4\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=0,15mol\\n_{Al_2\left(SO_4\right)_3}=0,3\cdot0,04=0,012\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,15}{3}>\dfrac{0,012}{1}\) \(\Rightarrow\) Ba(OH)2 dư, Al2(SO4)3 phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{Al\left(OH\right)_3}=0,024mol\\n_{BaSO_4}=0,036mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al\left(OH\right)_3}=0,024\cdot78=1,872\left(g\right)\\m_{BaSO_4}=0,036\cdot233=8,388\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{kếttủa}=1,872+8,388=10,26\left(g\right)\)
