a.b.\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{8}{160}=0,05mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,05 0,15 0,1 ( mol )
\(m_{Fe}=n.M=0,1.56=5,6g\)
\(V_{H_2}=n.22,4=0,15.22,4=3,36l\)
c.\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{1,12}{22,4}=0,05mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,1 > 0,05 ( mol )
0,075 0,05 0,025 ( mol )
\(m_{Fe_3O_4}=n.M=0,025.232=5,8g\)
\(m_{Fe\left(dư\right)}=n.M=\left(0,1-0,075\right).56=1,4g\)
Fe2O3+3H2-to>2Fe+3H2O
0,05------0,15-----0,1
n Fe2O3=\(\dfrac{8}{160}=0,05mol\)
=>VH2=0,15.22,4=3,36l
=>m Fe=0,1.56=5,6g
3Fe+2O2-to>Fe3O4
0,05--------0,025
n O2=\(\dfrac{1,12}{22,4}\)=0,05 mol
=> Fe dư
=>m Fe3O4=0,025.232=5,8g
nFe2O3 = 8:160 = 0,05 (mol)
pthh : Fe2O3 + 3H2 -t-->2 Fe +3H2O
0,05 -------> 0,15--->0,1--->0,15(mol)
mFe= 0,1 . 56=5,6 (g)
VH2 = 0,15.22,4=3,36(l)
nO2=1,12 :22,4 =0,05 (mol)
pthh : 3Fe + 2O2 -t--> Fe3O4
0,05 ----> 0,025(mol)
=> mFe3O4= 0,025 . 232 =5,8 ( g)