\(n_{HNO_3}=1,2.0,45=0,54\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{FeO}=x\\n_{CuO}=y\\n_{MO}=z\end{matrix}\right.\) ( mol ) \(\Rightarrow m_{hh}=72x+80y+\left(M+16\right)z=11,52\left(g\right)\left(\cdot\right)\)
`@`TH1: M bị H2 khử
\(FeO+H_2\rightarrow\left(t^o\right)Fe+H_2O\) (1)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\) (2)
\(MO+H_2\rightarrow\left(t^o\right)M+H_2O\) (3)
\(Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\) (4)
\(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\) (5)
\(3M+8HNO_3\rightarrow3M\left(NO_3\right)_2+2NO+4H_2O\) (6)
\(MO+2HNO_3\rightarrow M\left(NO_3\right)_2+H_2O\) (7)
Theo pt \(\left(1\right);\left(2\right);\left(3\right);\left(4\right);\left(5\right);\left(6\right)\):
\(\Rightarrow n_{HNO_3}=4x+\dfrac{8}{3}y+\dfrac{8}{3}z=0,54\)
Ta có: \(x:y:z=5:3:1\)
`=>` không có giá trị thỏa mãn
`@`TH2: M không bị khử bởi H2
Theo pt: \(\left(1\right);\left(2\right);\left(4\right);\left(5\right);\left(7\right)\) :
\(n_{HNO_3}=4x+\dfrac{8}{3}y+2z=0,54\)
Ta có: \(x:y:z=5:3:1\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,09\\y=0,054\\z=0,018\end{matrix}\right.\)
\(\left(\cdot\right)\Leftrightarrow72.0,09+80.0,054+\left(M+16\right).0,018=11,52\)
\(\Rightarrow M=24\) \((g/mol)\)
`=>` \(M\) là Magie ( Mg )
Theo pt: \(\left(1\right);\left(2\right);\left(4\right);\left(5\right)\) :
\(n_{NO}=x+2y=0,09+2.0,054=0,198\left(mol\right)\)
\(\Rightarrow V_{NO}=0,198.22,4=4,4352\left(mol\right)\)
