\(n_{FeCl_3}=0,5.1=0,5mol\)
FeCl3+3NaOH\(\rightarrow\)Fe(OH)3+3NaCl
2Fe(OH)3\(\overset{t^0}{\rightarrow}\)Fe2O3+3H2O
\(n_{NaOH}=3n_{FeCl_3}=1,5mol\)
\(V_{NaOH}=\dfrac{n}{C_M}=\dfrac{1,5}{0,5}=3l\)
\(C_{M_{NaCl}}=\dfrac{n}{v}=\dfrac{1,5}{0,05+3}=0,225M\)
\(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=\dfrac{1}{2}n_{FeCl_3}=0,25mol\)
M=\(m_{Fe_2O_3}=0,25.160=40g\)