Bài 4 :\(n_{Fe_3O_4}=0,06\left(mol\right);n_{N_xO_y}=0,02\left(mol\right)\)
\(Fe^{\text{8/3}}_3\rightarrow3Fe^{3+}+1e\) \(xN^{+5}+\left(5x-2y\right)\rightarrow N^{\text{2y/x}}_x\)
Bảo toàn e ta có : \(0,06.1=0,02.\left(5x-2y\right)\)
=> \(5x-2y=3\)
=> x=1 ; y=1
=> Khí cần tìm là NO
Bài 3:
\(Al\rightarrow Al^{3+}+3e\) \(10H^++2NO_3^-+8e\rightarrow N_2O+5H_2O\)
\(4H^++NO_3^-+3e\rightarrow NO+2H_2O\)
Bảo toàn e => \(n_{Al}=\dfrac{n_{N_2O}.8+n_{NO}.3}{3}=0,05\left(mol\right)\)
=> \(m_{Al}=0,05.27=1,35\left(g\right)\)
Bài 1 : \(Al\rightarrow Al^{3+}+3e\) \(10H^++2NO_3^-+8e\rightarrow N_2O+5H_2O\)
\(4H^++NO_3^-+3e\rightarrow NO+2H_2O\)
Bảo toàn e => \(n_{Al}=\dfrac{n_{N_2O}.8+n_{NO}.3}{3}=0,68\left(mol\right)\)
=> \(m_{Al}=0,68.27=18,36\left(g\right)\)
Bài 2 : \(Fe\rightarrow Fe^{3+}+3e\) \(2H^++NO_3^-+1e\rightarrow NO_2+H_2O\)
\(4H^++NO_3^-+3e\rightarrow NO+2H_2O\)
Bảo toàn e : \(n_{Fe}.3=n_{NO_2}.1+n_{NO}.3\Rightarrow n_{Fe}=0,03\left(mol\right)\)
=> \(m_{Fe}=0,03.56=1,68\left(g\right)\)
