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Dεερακ Ȿιηɠꜧ 4 years ago
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> $$ \mathbb{ANSWER\ IS \ COMPLETED} $$
Dεερακ Ȿιηɠꜧ 4 years ago
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>
Dεερακ Ȿιηɠꜧ 4 years ago
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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