Đặt nFe=x(mol),nAl=y(mol)(x,y>0) nH2(dktc)=22,40,56=0,025(mol)
Ta có hệ: {56x+27y=0,83x+23y=0,025⇔{x=0,01y=0,01 mFe=0,01.56=0,56(g),mAl=0,01.27=0,27(g) %mFe=0,830,56.100=67,47(%),%mAl=32,53(%)
Đặt nFe=x(mol),nAl=y(mol)(x,y>0) nH2(dktc)=22,40,56=0,025(mol) View attachment 221996
Ta có hệ: {56x+27y=0,83x+23y=0,025⇔{x=0,01y=0,01 mFe=0,01.56=0,56(g),mAl=0,01.27=0,27(g) %mFe=0,830,56.100=67,47(%),%mAl=32,53(%)