n(Fe)=14/56=0.25mol; n(NO)=4.48/22.4=0.2mol;
设生成Fe(NO3)2为xmol,则生成Fe(NO3)3为(0.25-x)mol;
根据电子转移守恒得2x+3(0.25-x)= 0.2(5-2)
n(NO3-)= 2x+3(0.25-x)+ 0.2 =0.2(5-2)+ 0.2 = 0.8mol
c(HNO3)=0.8/0.5 = 1.6(mol/L)
选B.
n(Fe)=14/56=0.25mol; n(NO)=4.48/22.4=0.2mol;
设生成Fe(NO3)2为xmol,则生成Fe(NO3)3为(0.25-x)mol;
根据电子转移守恒得2x+3(0.25-x)= 0.2(5-2)
n(NO3-)= 2x+3(0.25-x)+ 0.2 =0.2(5-2)+ 0.2 = 0.8mol
c(HNO3)=0.8/0.5 = 1.6(mol/L)
选B.