Ta có:
MX=22,6.2=45,2 mà X chắc chắn chứa CO2 (44) nên khí còn lại là NO2 (46).
Gọi \(\left\{{}\begin{matrix}n_{CO2}:a\left(mol\right)\\n_{NO2}:b\left(mol\right)\end{matrix}\right.\)
\(n_X=a+b=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(\Rightarrow m_X=44a+46b=45,2.0,15\Rightarrow\left\{{}\begin{matrix}a=0,06\\b=0,09\end{matrix}\right.\)
\(\Rightarrow n_{FeCO3}=n_{CO2}=0,06\left(mol\right)\)
\(Fe_3O_4+10HNO_3\rightarrow3Fe\left(NO_3\right)_3+NO_3+5H_2O\)
\(FeCO_3+4HNO_3\rightarrow Fe\left(NO_3\right)_3+CO_2+NO_2+H_2O\)
\(n_{NO2}=0,09\left(mol\right)\Rightarrow n_{Fe3O4}=0,09-0,06=0,03\left(mol\right)\)
\(\Rightarrow m=0,03.232+0,06.116=13,92\left(g\right)\)
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