\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a,
\(n_{H2}=\frac{7,28}{22,4}=0,325\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Al}:a\left(mol\right)\\n_{Fe}:b\left(mol\right)\end{matrix}\right.\)
Giải hệ PT:
\(\left\{{}\begin{matrix}27a+56b=6,8\\1,5a+b=0,325\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,025\end{matrix}\right.\)
\(\Rightarrow\%m_{Al}=\frac{0,2.27}{6,8}.100\%=79,41\%\)
\(\Rightarrow\%m_{Fe}=100\%-79,41\%=20,59\%\)
b,Ta có:
\(\left\{{}\begin{matrix}n_{Al}=0,3\left(mol\right)\\n_{Fe}=0,375\left(mol\right)\end{matrix}\right.\)
\(2Al+6H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(\Rightarrow n_{SO2}=0,45+0,05625=0,50625\left(mol\right)\)
\(\Rightarrow V_{SO2}=0,50625.22,4=11,34\left(l\right)\)
