\(n_{Fe_2O_3}=0,1mol\)
Fe2O3+6HCl\(\rightarrow\)2FeCl3+3H2O
\(n_{HCl}=6n_{Fe_2O_3}=0,6mol\)
\(n_{FeCl_3}=2.0,1=0,2mol\)
\(m_{dd_{HCl}}=\dfrac{0,6.36,5.100}{7,3}=300g\)
mdd=16+300=316g
\(C\%_{FeCl_3}=\dfrac{0,2.162,5.100}{316}\approx10,3\%\)
Fe(OH)3+3NaOH\(\rightarrow\)Fe(OH)3+3NaCl
2Fe(OH)3\(\rightarrow\)Fe2O3+3H2O
Số mol Fe2O3=0,2:2=0,1mol
m=0,1.160=16g