a) Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\Rightarrow27a+24b=3,75\left(1\right)\)
\(n_{H_2}=\dfrac{3,92}{22,4}=0,175\left(mol\right)\)
PTHH:
2Al + 6HCl ---> 2AlCl3 + 3H2
a---->3a-------->a-------->1,5a
Mg + 2HCl ---> MgCl2 + H2
b---->2b-------->b------->b
=> 1,5a + b = 0,175 (2)
Từ (1), (2) => a = 0,05; b = 0,1
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,05.27}{3,75}.100\%=36\%\\\%m_{Mg}=100\%-36\%=64\%\end{matrix}\right.\)
b) \(m_{muối}=0,05.133,5+0,1.95=16,175\left(g\right)\)
c) \(n_{HCl\left(pư\right)}=0,05.3+0,1.2=0,35\left(mol\right)\)
=> \(n_{HCl\left(bđ\right)}=0,35.\left(100+20\right)\%=0,42\left(mol\right)\)
=> \(V_{ddHCl}=\dfrac{0,42}{1}=0,42\left(l\right)=420\left(ml\right)\)
