\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=0,2.0,125=0,025\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,1 0,05 ( mol )
\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
0,025 0,05 ( mol )
\(m_{Fe}=0,05.56=2,8\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{2,8}{9,2}.100=30,43\%\\\%m_{Cu}=100\%-30,43\%=69,57\%\end{matrix}\right.\)
\(x=C_{M_{HCl}}=\dfrac{0,1+0,05}{0,3}=0,5\left(M\right)\)
a) \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right);n_{Ba\left(OH\right)_2}=0,125.0,2=0,025\left(mol\right)\)
PTHH:
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05<-0,1<--------------0,05
\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
0,025------->0,05
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,05.56}{9,2}.100\%=30,43\%\\\%m_{Cu}=100\%-30,43\%=69,57\%\end{matrix}\right.\)
b) \(x=C_{M\left(HCl\right)}=\dfrac{0,05+0,1}{0,3}=0,5M\)
