PTHH : \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H2}=\frac{V_{H2}}{22,4}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
- Theo PTHH : \(n_{Fe}=n_{H2}=0,15\left(mol\right)\)
=> \(m_{Fe}=n.M=0,15.56=8,4\left(g\right)\)
=> \(\%Fe=\frac{m_{Fe}}{m_{hh}}.100\%=\frac{8,4}{10}.100\%=84\%\)
Mà \(m_{hh}=m_{Fe}+m_{Cu}=10\left(g\right)\)
=> \(m_{Cu}=10-8,4=1,6\left(g\right)\)
=> \(\%Cu=\frac{m_{Cu}}{m_{hh}}.100\%=\frac{1,6}{10}.100\%=16\%\)
nH2= 3,36/22,4=0,15 mol
Cu+HCl => ko phản ứng
Fe + 2HCl => FeCl2 + H2
0,15 < ------------------0,15
=> mFe = 0,15.56 = 8,4(g)
=>%Fe = 84%
%Cu = 100% - %Fe = 100% - 84% = 16%
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