Gọi khối lượng của Al và Fe lần lược là x, y thì ta có hệ
\(\left\{{}\begin{matrix}x+y=27,6\\\dfrac{x}{y}=\dfrac{2,7}{4,2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10,8\\y=16,8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{10,8}{27}=0,4\\n_{Fe}=\dfrac{16,8}{56}=0,3\end{matrix}\right.\)
PTHH
\(4Al\left(0,4\right)+3H_2SO_4\left(0,3\right)\rightarrow2Al_2\left(SO_4\right)_3+3H_2\left(0,3\right)\)
\(Fe\left(0,3\right)+H_2SO_4\left(0,3\right)\rightarrow FeSO_4+H_2\left(0,3\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,3+0,3=0,6\\n_{H_2}=0,3+0,3=0,6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%\left(H_2SO_4\right)=\dfrac{0,6.98}{200}.100\%=29,4\%\\V_{H_2}=0,6.22,4=13,44\left(l\right)\end{matrix}\right.\)