\(n_k=n_{H_2}=0,125\left(mol\right)\)
a,b, \(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
.............0,125...0,125....................0,125...
\(\Rightarrow m_{Fe}=7\left(g\right)\)
Do Cu không phản ứng với H2SO4 .
\(\Rightarrow m_{Cu}=m_{hh}-m_{Fe}=10-7=3\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%Fe=70\%\\\%Cu=30\%\end{matrix}\right.\)
c, Có : \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=206,75\left(g\right)\)
\(\Rightarrow C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%\approx5,925\%\)