n(Fe)=
5.6g
56g/mol =0.1mol,生成NO的质量为5.6g-3.2g=2.4g,n(NO)=
2.4g
30g/mol =0.08mol,
设Fe 2+和Fe 3+物质的量分别为x、y,则
x+y=0.1
2x+3y=0.08×(5-2) ,
解得x=0.06mol,y=0.04mol,
所以所得溶液中Fe 2+和Fe 3+物质的量之比为0.06mol:0.04mol=3:2,
故选C.
n(Fe)=
5.6g
56g/mol =0.1mol,生成NO的质量为5.6g-3.2g=2.4g,n(NO)=
2.4g
30g/mol =0.08mol,
设Fe 2+和Fe 3+物质的量分别为x、y,则
x+y=0.1
2x+3y=0.08×(5-2) ,
解得x=0.06mol,y=0.04mol,
所以所得溶液中Fe 2+和Fe 3+物质的量之比为0.06mol:0.04mol=3:2,
故选C.