PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (1)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (2)
a) Ta có: \(n_{H_2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
Gọi số mol của Fe là \(a\) \(\Rightarrow n_{H_2\left(1\right)}=a\)
Gọi số mol của Al là \(b\) \(\Rightarrow n_{H_2\left(2\right)}=\frac{3}{2}b\)
Ta lập được hệ phương trình;
\(\left\{{}\begin{matrix}a+\frac{3}{2}b=0,6\\56a+27b=22,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,3mol\\n_{Al}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,3\cdot56}{22,2}\cdot100\approx75,68\%\\\%m_{Al}=24,32\%\end{matrix}\right.\)
b)Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(1\right)}=n_{Fe}=0,3mol\\n_{H_2SO_4}=\frac{3}{2}n_{Al}=0,3mol\end{matrix}\right.\)
\(\Rightarrow n_{H_2SO_4}=0,6mol\) \(\Rightarrow m_{ddH_2SO_4}=\frac{0,6\cdot98}{4,9\%}=1200\left(g\right)\)
c) Ta có: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,3mol\\n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,3\cdot152=45,6\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{H_2}=0,6\cdot2=1,2\left(g\right)\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}=1221\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\frac{45,6}{1221}\cdot100\approx3,73\%\\C\%_{Al_2\left(SO_4\right)_3}=\frac{34,2}{1221}\cdot100\approx2,8\%\end{matrix}\right.\)
