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two oxides oa a metal contain 72.4% and 70% of metal respectively? if formula of 2nd oxide isM2O3, find that of the first
Dear student For Oxygen, we can calculate as w(%) = 100 -70 = 30But Ar(O) = 16Based on the law of proportion - [% / Ar], so according to it,= 70/x : 30/16 = 2:3= 70/x : 1.875 = 2:3, where x = 56 (Fe) So, if w(O) = 100 – 72.4 = 27.6 (%) =72.4/56 : 27.6/16 = 1.2928 : 1.725 = 1 : 1.334 = 3 : 4 The answer should be - M3O4 (Fe3O4) RegardsArun
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