\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a,
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
Gọi số mol của Al là x, số mol của Fe là y
\(\left\{{}\begin{matrix}27x+56y=11,1\\1,5a+b=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
\(\%m_{Al}=\frac{0,1.27}{11,1}.100\%=24,32\%\)
\(\%m_{Fe}=100\%-24,32\%=75,68\%\)
b,\(n_{H2SO4}=\frac{245.25\%}{98}=0,625\left(mol\right)\)
\(\Rightarrow n_{H2SO4.trong.Y}=0,625-0,3=0,325\left(mol\right)\)
\(m_{dd\left(spu\right)}=11,1+245-0,3.2=255,5\left(g\right)\)
\(n_{Al2\left(SO4\right)3}=0,05\left(mol\right)\)
\(n_{FeSO4}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{HeSO4}=12,47\%\\C\%_{Al2\left(SO4\right)3}=6,2\%\\C\%_{FeSO4}=8,92\%\end{matrix}\right.\)
