PTHH: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
Mol: x 3x x
\(m_{Fe_2O_3}=160x\left(g\right)\)
\(m_{H_2SO_4}=3x.98=294x\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{294x.100\%}{10\%}=2940x\left(g\right)\)
⇒mdd sau pứ = 160x+2940x = 3100x
\(m_{Fe_2\left(SO_4\right)_3}=400x\)
\(\Rightarrow C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{400x.100\%}{3100x}=12,9\%\)