Question

Answer 1

\(n_{HCl}=0,4.1=0,4\left(mol\right)\\ n_{Zn}=a,n_{Fe}=b\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}2a+2b=0,4\\65a+56b=12,1\end{matrix}\right.\\ \Rightarrow a=b=0,1\\ n_{ZnCl_2}=n_{Zn}=n_{Fe}=n_{FeCl_2}=0,1mol\\ m_{muối}=0,1\left(136+127\right)=26,3\left(g\right)\)

Answer 2

\(n_{Cl\left(trongHCl\right)}=n_{HCl}=0,4.1=0,4\left(mol\right)\\ m_{muối}=m_{hh\left(Zn,Fe\right)}+m_{Cl\left(trongHCl\right)}=12,1+0,4.35,5=26,3\left(g\right)\)