a) \(Mg+2HCl-->MgCl2+H2\)
x--------------------------------------------x(mol)
\(2Al+6HCl-->2AlCl3+3H2\)
y------------------------------------1,5y(mol)
\(n_{H2}=\frac{1,568}{22,4}=0,07\left(mol\right)\)
\(m_{Al}+m_{Mg}=3,42-1,92=1,5\left(g\right)\)
Theo bài ra ta có hpt
\(\left\{{}\begin{matrix}24x+17y=1,5\\x+1,5y=0,07\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,02\end{matrix}\right.\)
\(\%m_{Mg}=\frac{0,04.24}{3,42}.100\%=28,07\%\)
\(\%m_{Al}=\frac{0,02.27}{3,42}.100\%=15,79\%\)
\(\%m_{Cu}=100-15,79-28,07=56,14\%\%\)
b) \(n_{HCl}=2n_{H2}=0,14\left(mol\right)\)
\(m_{HCl}=0,14.36,5=5,11\left(g\right)\)
\(m=m_{ddHCl}=\frac{5,11.100}{10}=51,1\left(g\right)\)
