nCO=896/(1000*22.4)=0.04(mol)
=> nCO(phản ứng)=0.04*80/100=0.032(mol)
Fe3O4 + 4CO --> 3Fe + 4CO2
0.008....0.032......0.021.....0.032(mol)
mFe3O4(Phản ứng)=0.008*232=1.856(g)
mFe3O4(trong X)=1.856*100/80=2.32(g)
mX=2.32*100/95=2.442(g)
ΣnB=nCO2+nCO(dư)=0.032+(0.04-0.032)=0.04(mol)
VB=0.04*22.4=0.896(l)
mA=mFe+mFe3O4(dư)+mTạp chất
=0.032*56+(2.32-1.856)+2.442*5/100=2.3781(g)
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