a)\(Fe+2HCl-->FeCl2+H2\)
\(Fe2O3+6HCl-->2FeCl3+3H2O\)
b)\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{Fe}=n_{H2}=0,2\left(mol\right)\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(\%m_{Fe}=\frac{11,2}{27,2}.100\%=41,18\%\)
\(\%m_{Fe2O3}=100-41,18=58,82\%\)
c)\(n_{HCl\left(1\right)}=2n_{H2}=0,4\left(mol\right)\)
\(n_{Fe2O3}=\frac{27,2-11,2}{160}=0,1\left(mol\right)\)
\(n_{HCl}=6n_{Fe2O3}=0,6\left(mol\right)\)
\(\sum n_{HCl}=0,1+0,6=0,7\left(mol\right)\)
\(V_{HCl}=\frac{0,7}{2,5}=0,28\left(l\right)=280ml\)
\(m_{ddHCl}=280.1,1=308\left(g\right)\)
m dd sau pư = \(m_{hh}+m_{ddHCl}-m_{H2}=27,2+308-0,4=334,8\left(g\right)\)
\(n_{FeCl2}=n_{H2}=0,2\left(mol\right)\)
\(C\%_{FeCl2}=\frac{0,2.127}{334,8}.100\%=7,59\%\)
\(n_{FeCl3}=2n_{Fe2O3}=0,2\left(mol\right)\)
\(C\%_{FeCl3}=\frac{0,2.162,5}{334,8}.100\%=9,7\%\)
