a, Ta có: \(n_{Fe}=\dfrac{1,26}{56}=0,0225\left(mol\right)\)
\(n_{CO_2}=\dfrac{1,32}{44}=0,03\left(mol\right)\)
\(CO+O_{\left(trongoxit\right)}\rightarrow CO_2\)
_____0,03_________0,03 (mol)
\(\Rightarrow n_{Fe}:n_O=0,0225:0,03=3:4\)
Vậy: CTHH của oxit đó là Fe3O4.
b, Ta có: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,0075\left(mol\right)\)
PT: \(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)
____0,0075__0,06 (mol)
\(\Rightarrow V_{HCl}=\dfrac{0,06}{1,5}=0,04\left(l\right)=40\left(ml\right)\)