a. \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{400.9,8\%}{98}=0,4\left(mol\right)\)
PTHH : Fe2O3 + 3H2SO4 ---to---> Fe2(SO4)3 + 3H2O
\(\dfrac{0.4}{3}\) 0,4 \(\dfrac{0.4}{3}\)
Ta thấy : \(\dfrac{0.2}{1}>\dfrac{0.4}{3}\) => Fe2O3 dư
b. Dung dịch sau phản ứng gồm Fe2O3 dư , Fe2(SO4)3
\(m_{Fe_2O_3\left(dư\right)}=\left(0,2-\dfrac{0.4}{3}\right).160=\dfrac{32}{3}\left(g\right)\)
\(m_{dd}=32+400=432\left(g\right)\)
\(C\%_{dd_{Fe_2O_3}}=\dfrac{32}{\dfrac{3}{432}}.100\%=2,46\%\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{\left(\dfrac{0.4}{3}.256\right)}{432}.100\%=7,9\%\)
\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(m_{H_2SO_4}=\dfrac{400.9,8\%}{100\%}=39,2\left(g\right)\) \(\Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
PTHH:
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
2/15 0,4 2/15 0,4 (mol)
LTL: \(\dfrac{0,2}{1}>\dfrac{0,4}{3}\) => Fe2O3 dư , H2SO4 đủ
\(m_{Fe_2\left(SO_4\right)_3}=\dfrac{2}{15}.400=\dfrac{160}{3}\left(g\right)\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{\dfrac{160}{3}.100}{32+400}=12,35\%\)