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Answer» $$CH_{4} + O_{2} {\\in} \\atop {∆} $$<br>The iron oxide has $$69.9%$$ iron and\xa0$$30.1%$$ dioxygen by mass.Thus, $$100 g$$ of iron oxide contains $$69.9 g$$ iron and $$30.1 g$$ dioxygen.The number of moles of iron present in $$100 g$$ of iron oxide are\xa0$$\\frac{69.9}{55.8}=1.25$$.The number of moles of dioxygen present in 100 g of iron oxide are\xa0$$\\frac{30.1}{32}=0.94$$.The ratio of the number of oxygen atoms to the number of carbon atoms present in one formula unit of iron oxide is\xa0$$\\frac{2×0.94}{1.25}=1.5:1=3:2$$.Hence, the formula of the iron oxide is\xa0$$Fe2O3$$<hr>$$ \\mathbb{ANSWER\\ IS \\ COMPLETED} $$<br>$$\\mathbb{MATHEMATICS \\ IS \\ BEAUTIFUL} $$<br>The iron oxide has $$69.9%$$ iron and\xa0$$30.1%$$ dioxygen by mass.Thus, $$100 g$$ of iron oxide contains $$69.9 g$$ iron and $$30.1 g$$ dioxygen.The number of moles of iron present in $$100 g$$ of iron oxide are\xa0$$\\frac{69.9\u200b}{55.8}=1.25$$.The number of moles of dioxygen present in 100 g of iron oxide are\xa0$$\\frac{30.1\u200b}{32}=0.94$$.The ratio of the number of oxygen atoms to the number of carbon atoms present in one formula unit of iron oxide is\xa0$$\\frac{2×0.94\u200b}{1.25}=1.5:1=3:2$$.Hence, the formula of the iron oxide is\xa0$$Fe2O3$$<hr><br>The iron oxide has $$69.9%$$ iron and\xa0$$30.1%$$ dioxygen by mass.Thus, $$100 g$$ of iron oxide contains $$69.9 g$$ iron and $$30.1 g$$ dioxygen.The number of moles of iron present in $$100 g$ of iron oxide are\xa0$$\\frac{69.9\u200b}{55.8}=1.25$$.The number of moles of dioxygen present in 100 g of iron oxide are\xa0$$\\frac{30.1\u200b}{32}=0.94$$.The ratio of the number of oxygen atoms to the number of carbon atoms present in one formula unit of iron oxide is\xa0$$\\frac{2×0.94\u200b}{1.25}=1.5:1=3:2$$.Hence, the formula of the iron oxide is\xa0$$Fe2O3$$<hr>


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