\(n_{BaCO_3}=\dfrac{27,58}{197}=0,14\left(mol\right)\)
PTHH: \(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)
0,14<---0,14
\(\Rightarrow n_{CO_2}=0,14\left(mol\right)\Rightarrow n_{O\left(oxit\right)}=0,14\left(mol\right)\)
\(n_{H_2}=\dfrac{2,352}{22,4}=0,105\left(mol\right)\)
PTHH: \(2M+2nHCl\rightarrow2MCl_n+nH_2\)
\(\dfrac{0,21}{n}\)<----------------------0,105
Ta có: \(m_{oxit}=m_M+m_O=\dfrac{0,21}{n}.M_M+0,14.16=8,12\left(g\right)\)
=> MM = 28n (g/mol)
Xét n = 2 thỏa mãn => MM = 56 (g/mol)
=> M là Fe
\(n_{Fe}=\dfrac{0,21}{2}=0,105\left(mol\right)\)
Xét nFe : nO = 0,105 : 0,14 = 3 : 4
=> CTHH: Fe3O4
