某Fe和Fe2O3d 混合物溶于1000ml的稀硝酸中,放出NO22.4L(标况）并余5.44gFe,向反应后的溶液中通入20.16LCl2

下面我是按已知量,步步求的未知量
n(NO)=22.4/22.4=1mol Cl2=20.16/22.4=0.9mol
Fe + 4HNO3 == Fe(NO3)3 + NO + 2H2O
1 4 1 1
1 4 1 1
2Fe2+ +Cl2 == 2Fe3+ +2Cl-
2 1
1.8 0.9
Fe + 2Fe(NO3)3 == 3Fe(NO3)2
1 2 3
0.6 1.2 1.8
Fe2O3 + 6HNO3 == 2Fe(NO3)3+ 3H2O
1 6 2
0.1 0.6 0.2
n(HNO3)=0,6+4=4.6 C=n/v=4.6/1=4.6mol/L
m(Fe)=(0.6+1)*56+5.44=95.04
m(Fe2O3)=0.1*160=16
Fe%=95.04/(95.04+16)*%=85.6%