a)\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
b)\(n_{H_2}:\dfrac{22,4}{22,4}=1\left(mol\right)\)
Gọi x, y lần lượt là số mol Zn, Fe
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
1.............................................1(mol)
x..............................................x(mol)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
1...........................................1(mol)
y............................................y(mol)
Ta có:\(\left\{{}\begin{matrix}65x+56y=60,5\\x+y=1\end{matrix}\right.\)
=>x=0,5,y=0,5
\(m_{Zn}:0,5.65=32,5\left(g\right)\)
\(m_{Fe}:0,5.56=28\left(g\right)\)
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