\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\left(1\right)\)
Cu không pw
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{H_2}=n_{Fe}=0,3\left(mol\right)\)
\(m_{Fe}=0,3.56=16,8g\)
\(m_{Cu}=20-16,8=3,2g\)
b) Lỡ gửi
\(CuO+CO\rightarrow Cu+CO_2\)
\(n_{Cu}=\dfrac{3,2}{64}=0,05mol\)
\(n_{CuO}=n_{Cu}=0,05\left(mol\right)\)
\(m_{CuO}=0,05.80=4g\)
a)bổ sung
\(\%Fe=\dfrac{16,8}{20}.100=84\%\)
%Cu=100%-84%=16%