Jeżeli zawartość procentowa żelaza wynosi 70% to:
100% - 70% = 30% tlenu
FexOy
mFe=56u (x)
mO=16u (y)
56x/16y = 70/30
1680x = 1120y
3x = 2y
Wzór tlenku to Fe₂O₃
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Jeżeli zawartość procentowa żelaza wynosi 70% to:
100% - 70% = 30% tlenu
FexOy
mFe=56u (x)
mO=16u (y)
56x/16y = 70/30
1680x = 1120y
3x = 2y
Wzór tlenku to Fe₂O₃
mFe₂O₃=112u+48u=160u
mFe : mO = 112 : 48
mFe : mO = 7 : 3