\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,02___0,04___0,02____0,02
\(Fe_2O_3+6HCl\rightarrow2FeCl_2+3H_2O\)
0,03____0,18 _____0,06_________
\(n_{HCl}=\frac{110:7,3}{36,5}=0,22\left(mol\right)\)
\(\frac{n_{H2}}{0,44822,4}=0,02\left(mol\right)\)
\(\%m_{Fe}=\frac{0,2.56}{0,02.56+0,03.160}.100\%=18,92\%\)
\(\%m_{Fe2O3}=100\%-18,92\%=81,08\%\)
\(\rightarrow m_{FeCl_2}=0,08.\left(56+71\right)=10,16\left(g\right)\)