\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PT: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
a, Theo PT: \(n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)
\(\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)
b, \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,5}=1,2\left(M\right)\)