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9:8
mAl2O3= 2 x mAl + 3 x mO = 2 x 27u + 3 x 16u = 54u + 48u = 102u
x- skład % glinu w tym związku
54u = x
102u = 100%
102x = 54 x 100%
102x = 5400%/:102
x = ok.53% glinu
100% - 53% = 47% tlenu