Ta có:
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{Cl2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
Gọi a,b lần lượt là số mol Fe,Zn
\(PTHH:TN_1:Fe+2HCl\rightarrow FeCl_2+H_2\)
_____________a____________a ________________
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b _____________b_________
\(TH_2:2Fe+3Cl_2\rightarrow2FeCl_3\)
________a___________1,5a
\(Zn+Cl_2\rightarrow ZnCl_2\)
b_______b__________
Giải hệ PT:
\(\left\{{}\begin{matrix}a+b=0,2\\1,5a+b=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow m=m_{Fe}+m_{Zn}=0,1.56+0,1.65=12,1\left(g\right)\)
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