Câu 30.
1.\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(m_{Fe}=0,2.56=11,2g\)
\(m_{chất.rắn}=m_{Cu}=3,2g\)
\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{11,2}{11,2+3,2}.100=77,78\%\\\%m_{Cu}=100\%-77,78\%=22,22\%\end{matrix}\right.\)
\(m_{hh}=11,2+3,2=14,4g\)
2.
\(2Fe+6H_2SO_4\left(đ\right)\rightarrow\left(t^o\right)Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
0,2 0,6 0,1 0,3 ( mol )
\(Cu+2H_2SO_4\left(đ\right)\rightarrow\left(t^o\right)CuSO_4+SO_2+2H_2O\)
0,05 0,1 0,05 0,05 ( mol )
\(V_{SO_2}=\left(0,3+0,05\right).22,4=7,84l\)
\(m_{H_2SO_4}=\dfrac{\left(0,6+0,1\right).98}{98\%}=70g\)
\(m_{ddspứ}=14,4+70-\left(0,3+0,05\right).64=62g\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{62}.100=64,51\%\)
\(C\%_{CuSO_4}=\dfrac{0,05.160}{62}.100=12,9\%\)
PTHH: \(\left\{{}\begin{matrix}Fe+2HCl\rightarrow FeCl_2+H_2\\Cu+2HCl\rightarrow CuCl_2+H_2\end{matrix}\right.\)
\(n_{H_2}=\dfrac{V_{H_2\left(đktc\right)}}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PTHH: \(n_{H_2}=n_{Fe}=n_{Cu}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=n.M=0,2.56=11,2\left(g\right)\\m_{Cu}=n.M=0,2.64=12,8\left(g\right)\end{matrix}\right.\)
\(a=m_{Fe}+m_{Cu}=11,2+12,8=24\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%Fe=\dfrac{11,2}{24}.100\approx46,7\%\\\%Cu=\dfrac{12,8}{24}.100\approx53,3\%\end{matrix}\right.\)
