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4,2g + xg = 5,8g
x = 1,6g tlenu
mFe : mO = 4,2 : 1,6
mFe : mO = 21 : 8
FexOy
mFe=56u (x)
mO=16u (y)
56x/16y = 21/8
56x*8 = 16y*21
448x = 336y
1,33x = y
4x = 3y
x : y = 3 : 4
Wzór tlenku:Fe₃O₄