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2014-05-25T15:32:21+03:00
28,57 g Fe2Ox................20 g Fe
112+16x g Fe2Ox...........2*56 g Fe
(112+16x)*20=2*56*28,57
2240+320x=3200=>32x=960=>x=3
oxidul Fe2O3 = oxidul de fier III
160 g Fe2O3......2*56 g Fe
x g FeO..........20 g Fe     x = 28,57 g Fe2O3 - adevarat!!
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