\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
Mol: 0,3 0,9 0,3
\(m_{H_2SO_4}=0,9.98=88,2\left(g\right)\)
\(m_{ddH_2SO_4}=\dfrac{88,2}{10}.100=882\left(g\right)\)
mdd sau pứ = 48+882 = 930 (g)
\(C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,3.400.100\%}{930}=30\%\)
\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ 0,3........0,9.............0,3............0,9\left(mol\right)\\ m=m_{ddH_2SO_4}=\dfrac{0,9.98.100}{10}=882\left(g\right)\\ m_{ddsau}=48+882=930\left(g\right)\\ C\%_{ddsau}=C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,3.400}{930}.100\approx12,903\%\)
