\(m_{HCl}=52.14\cdot1.05\cdot10\%=5.4747\left(g\right)\)
\(n_{HCl}=0.15\left(mol\right)\)
\(Fe_aO_b+2bHCl\rightarrow aFeCl_{\dfrac{2b}{a}}+bH_2\)
\(\dfrac{0.15}{2b}..........0.15\)
\(M=\dfrac{4}{\dfrac{0.15}{2b}}=\dfrac{160}{3}b\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow56a+16b=\dfrac{160}{3}b\)
\(\Leftrightarrow56a=\dfrac{112}{3}b\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{2}{3}\)
\(Fe_2O_3\)
