\(\text{3Cu+8HNO3=3Cu(NO3)2+2NO+4H2O}\)
\(\text{Fe2O3+6HNO3=2Fe(NO3)3+3H2O}\)
nNO=3,36/22,4=0,15mol
=>nCu=0,15.3:2=0,225mol
=>mCu=0,225.64=14,4g
=>mFe2O3=30,4-14,4=16g
=>nFe2O3=16/160=0,1mol
\(\text{Spu thu đc 0,225mol Cu(NO3)2; 0,2mol Fe(NO3)3}\)
\(\text{Cu(NO3)2=CuO+2NO2+1/2O2}\)
\(\text{0,225mol Cu(NO3)2 sinh ra 0,45mol NO2, 0,1125mol O2}\)
\(\text{2Fe(NO3)3=Fe2O3+6NO2+3/2O2}\)
\(\text{0,2 mol Fe(NO3)3 sinh ra 0,6mol NO2, 0,15mol O2}\)
Vậy sau khi nhiệt phân muối, thu đc 0,45+0,1125+0,6+0,15=1,3125mol
=> V khí= \(\text{V=1,3125.22,4=29,4l}\)
28HNO3 | + | 3Fe3O4 | → | 14H2O | + | NO | + | 9Fe(NO3)3 |
b) n NO=3,36/33,4=0,15(mol)
Theo pthh
n Fe2O3=3n NO=0,45(mol)
m Fe2O3=0,45.160=72(g)
-->Đề sai