1) \(Fe_2O_3+3H_2-t^0>2Fe+3H_2O\)
\(2Fe+3Cl_2-t^0->2FeCl_3\)
\(FeCl_3+3NaOH->Fe\left(OH\right)3+3NaCl\)
\(n_{NaOH}=0,5\left(mol\right)\) \(n_{H2SO4}=0,27\left(mol\right)\)
\(2NaOH+H_2SO_4->Na_2SO_4+2H_2O\)
0,5....................0,25...............0,25
\(\dfrac{0,5}{2}< \dfrac{0,27}{1}\) => H2SO4 dư
\(m_{Na_2SO_4}=0,25.142=35,5\left(g\right)\)