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All the oxygen in 0.5434g sample of pure oxide of iron is removed by reduction in a stream of H_(2). The loss in weight is 0.1210g. What is the formula of the oxide of iron ? (At. Mass of Fe = 56) |
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Answer» SOLUTION :Iron oxide = 0.5434 g Oxygen lost as `H_(2)O=0.1212g` `therefore"Iron "=0.5434-0.1210=0.4224g` `therefore""%" of FE"=(0.4224)/(0.5434)xx100=77.73%` `%" of O=100-77.73=22.27%`  Hence, formula of iron oxide = FEO. |
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