Mg+2HCl--->MgCl2+H2
x-----2x
2Al+6HCl---->2AlCl3+3H2
y--------3y
n\(_{HCl}=0,2.2=0,4\left(mol\right)\)
Theo bài ra ta có pt
\(\left\{{}\begin{matrix}24x+27y=3,9\\2x+3y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
----->m\(_{Mg}=0,05.24=1,2\left(g\right)\)
\(\text{Mg + 2HCl ---> MgCL2 + H2}\)
\(\text{2Al + 6HCl ----> 2AlCL3 + 3H2}\)
Ta có :
mhh= 24a + 27b =3,9
nHCl = 2a + 3b =0,4
Giải 2 PT trên
=> a=0,05 ;b=0,1
\(\text{-> m Mg=1,2 g}\)
