a)\(n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Gọi n Mg =x , n Zn =y
\(Mg+2HCl-->MgCl2+H2\)
x----------------------------------------x(mol)
\(Zn+2HCl-->ZnCl2+H2\)
y-------------------------------------y(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}24x+65y=3,9\\x+y=0,1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,063\\y=0,037\end{matrix}\right.\)
\(\%m_{Mg}=\frac{24,0,063}{3,9}.100\%=38,77\%\)
\(\%m_{Zn}=100-38,77=61,23\%\)
b)\(n_{HCl}=2n_{H2}=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)
Áp dụng ĐLBTKL
\(m_{muối}=m_{KL}+m_{HCl}-m_{H2}=3,9+7,3-0,2\)
=\(11\left(g\right)\)
