a) Fe2O3+3H2--->2Fe+3H2O
b) n H2=6,72/22.4=0,3(mol)
Theo pthh
n Fe2O3=1/3n H2=0,1(mol)
m Fe2O3=0,1.160=16(g)
c) 2H2+O2--->2H2O
n kk=19,6/22,4=0,875(mol)
n O2=0,875/5=0,175(mol)
lập tỉ lệ
n H2=0,3/2=0,15(mol)
n O2=0,175/1=0,175(mol)
--->O2 dư
Theo pthh2
n H2O=n H2=0,3(mol)
m H2O=0,3.18=5,4(g)