Ta có:
\(n_{Fe}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\Rightarrow n_{H2}=n_{Fe}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{8,4}{12}.100\%=70\%\\\%m_{Ag}=100\%-70\%=30\%\end{matrix}\right.\)
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