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