\(n_{H_2SO_4}=0,12\cdot1,5=0,18\left(mol\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,06 ← 0,18 → 0,06
\(\Rightarrow m_{Fe_2O_3}=0,06\cdot160=9,6\left(g\right)\)
\(\Rightarrow C_{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{0,06}{0,12}=0,5\left(M\right)\)
\(n_{H_2SO_4}=0,12.1,5=0,18\left(mol\right)\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{Fe_2O_3}=n_{Fe_2\left(SO_4\right)_3}=\dfrac{0,18}{3}=0,06\left(mol\right)\\ m_{Fe_2O_3}=160.0,06=9,6\left(g\right)\\ V_{ddsau}=V_{ddH_2SO_4}=0,12\left(l\right)\\ C_{MddFe_2\left(SO_4\right)_3}=\dfrac{0,06}{0,12}=0,5\left(M\right)\)
