\(\sum n_{H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ \Rightarrow\sum m_{H_2}=0,125\cdot2=0,25\left(g\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\sum n_{HCl}=2\sum n_{H_2}=0,25\left(mol\right)\\ \Rightarrow\sum m_{HCl}=0,25\cdot36,5=9,125\left(g\right)\\ \Rightarrow m_{\text{muối khan}}=m_{hh}+\sum m_{HCl}-\sum m_{H_2}=5,1+9,125-0,25=13,975\left(g\right)\)
Mg + 2HCl ---> MgCl2 + H2 (1)
x -------------> x (mol)
2Al + 6HCl ---> 2AlCl3 + 3H2 (2)
y -----------------> 3/2 x (mol)
=> \(\left\{{}\begin{matrix}24x+27y=5,1\\x+\dfrac{3}{2}y=\dfrac{2,8}{22,4}=0,125\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,475\left(mol\right)\\y=-\dfrac{7}{30}\left(mol\right)\end{matrix}\right.\)
kh bit mik sai hay j nhma nó kh ra =))) cậu check lại thử đề bài xemm
