a) nFe2O3=32/160=0,2(mol)
nFe =17,92/56=0,32(mol)
theo pthh :nFe2O3=1/2nFe=0,16(mol)
=>H =0,2/0,16 .100=62,5(%)
b) theo pthh : nCO2 =3/2nFe=0,48(mol)
=> Vco2=0,48.22,4=10,752(l)
a)
\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
nFe \(=\dfrac{17,92}{56}=0,32\left(mol\right)\)
PT
Fe2O3 + 3CO ---to---> 2Fe + 3CO2
0,2...........................................0,4 (mol)
=> H = \(\dfrac{0,32}{0,4}.100\%=80\%\)
b) VCO2 = \(\dfrac{22,4.0,32.3}{2}=10,752\left(l\right)\)