Question

How many grams of $F{e}_2{O}_3$ can be formed from the rusting of $446$ grams of $Fe$ according to the reaction: $4Fe+3O_2\to 2F{e}_2{O}_3$ and excess oxygen?

• A. $320g$
• B. $223g$
• C. $159g$
• D. $480g$
• E. $640g$

Answer 1

The correct option is E $640g$

The atomic masses of $Fe$ and $O$ are 55.8 g/mol and 16 g/mol respectively.

The number of moles is the ratio of mass to molar mass.

The number of moles of Fe $=\frac{446g}{55.8g/mol}=8$ moles.

8 moles of Fe will give $\frac{8\times 2}{4}=4$ moles of $F{e}_2{O}_3$

The molar mass of ($F{e}_2{O}_3$) = $2\left(55.8\right)+3\left(16\right)=159.6$ g/mol

The mass of $F{e}_2{O}_3$ obtained $=159.6\times 4=640$ g

Hence, option E is correct.