\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,1 < 0,4 ( mol )
0,1 0,3 0,1 ( mol )
\(m_{ddspứ}=16+200=216\left(g\right)\)
\(\left\{{}\begin{matrix}C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{216}.100=18,51\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,3\right).98}{216}.100=4,54\%\end{matrix}\right.\)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=\dfrac{200.19,6\%}{100\%}=39,2\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
LTL : \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\)
=> H2SO4 dư , Fe đủ
dd sau pứ gồm H2SO4 dư và Fe2(SO4)3
\(n_{H_2SO_{4\left(sau.pứ\right)}}=0,4-0,3=0,1\left(mol\right)\)
\(m_{H_2SO_{4\left(dư.sau.pứ\right)}}=0,1.98=9,8\left(g\right)\)
\(m_{Fe_2\left(SO_4\right)_3}=0,1.400=40\left(g\right)\)
\(m_{dd}=mFe_2O_3+m_{dd}H_2SO_4=16+200=216\left(g\right)\)
\(C\%_{ddH_2SO_{4\left(sau.pứ\right)}}=\dfrac{9,8.100}{216}=4,54\%\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{40.100}{216}=18,52\%\)
