Đáp án:

\[a) \, \begin{cases}\%m_{Al}=57,5\%\\\%m_{Mg}=42,5\%\end{cases}\\ b)\,\begin{cases} C\%_{MgSO_4(X)}=2,96\%\\ C\%_{Al_2{(SO_4)}_3(X)}=5,066\%\\ C\%_{H_2SO_4(X)}=2,9\%\end{cases}\]

Giải thích các bước giải:

\(2Al+3H_2SO_4\to Al_2{(SO_4)}_3+3H_2\\ Mg+H_2SO_4\to MgSO_4+H_2\\ n_{H_2}=\dfrac{1,568}{22,4}=0,07\text{(mol)}\\ n_{H_2SO_4}=\dfrac{100\times 9,8\%}{98}=0,1\text{(mol)}\)

\(a)\) Gọi số mol Al, Mg trước phản ứng lần lượt là $x;y$
\(\to \begin{cases}27x+24y=1,41\\ 1,5x+y=0,07\end{cases}\to \begin{cases}a=0,03\text{(mol)}\\b=0,025\text{(mol)}\end{cases}\to m_{Al}=27\times 0,03=0,81\text{(g)}\to \%m_{Al}=\dfrac{0,81}{1,41}\times 100\%=57,5\%\to \%m_{Mg}=100\%-57,5\%=42,5\%\)

\(b)\, n_{H_2SO_4\,\text{phản ứng}}=n_{H_2}=0,07\text{(mol)}\to n_{H_2SO_4\,\text{dư}}=0,1-0,07=0,03\text{(mol)}\to m_{H_2SO_4\,\text{dư}}=0,03\times 98=2,94\text{(g)}\)

\(n_{Al_2{(SO_4)}_3}=\dfrac 12\times n_{Al}=\dfrac 12\times 0,03=0,015\text{(mol)}\to m_{Al_2{(SO_4)}_3}=0,015\times 342=5,13\text{(g)}\\n _{MgSO_4}=n_{Mg}=0,025\text{(mol)}\to m_{MgSO_4}=0,025\times 120=3\text{(g)} \)

BTKL: 

\(m_X=1,41+100-0,07\cdot 2=101,27\ \text{gam}\)

\(\to\begin{cases} C\%_{MgSO_4(X)}=\dfrac 3{101,27}\times 100\%=2,96\%\\ C\%_{Al_2{(SO_4)}_3(X)}=\dfrac{5,13}{101,27}\times 100\%=5,066\%\\ C\%_{H_2SO_4(X)}=\dfrac{2,94}{101,27}\times 100\%=2,9\%\end{cases}\)