\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{294.20\%}{100\%}:98=0,6\left(mol\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,1------>0,3--------->0,1
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,6}{3}\Rightarrow\) H2SO4 dư
\(m_{H_2SO_{4.dư}}=\left(0,6-0,3\right).98=29,4\left(g\right)\)
b.
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400.100\%}{16+294}=12,9\%\)
\(C\%_{H_2SO_4}=\dfrac{29,4.100\%}{16+294}=9,48\%\)
\(a,n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\\ n_{H_2SO_4}=\dfrac{294.20}{100.98}=0,6mol\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ \rightarrow\dfrac{0,1}{1}< \dfrac{0,6}{3}\Rightarrow H_2SO_4.dư\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1mol\\ n_{H_2SO_4}=3n_{Fe_2O_3}=0,3mol\\ m_{H_2SO_4.dư}=\left(0,6-0,3\right).98=29,4g\\ b,C_{\%Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{16+294}\cdot100=12,9\%\\ C_{\%H_2SO_4.dư}=\dfrac{29,4}{16+294}\cdot100=9,48\%\%88\)
