Question

An iron nail rusts when exposed to oxygen. According to the following reaction, how many grams of iron(III) oxide will be formed upon the complete reaction of 31.1 grams of oxygen gas with excess iron?
iron(s) + oxygen(g) → iron(III) oxide(s)

_______ moles iron(III) oxide

Answer 1

Answer: 103.8 g of $Fe_2O_3$, 0.65 moles of

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} O_2=\frac{31.1}{32}=0.975moles$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $Fe$ is the excess reagent.

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3$

According to stoichiometry :

3 moles of $O_2$ produce =  2 moles of $Fe_2O_3$

Thus 0.975 moles of $O_2$ will produce=$\frac{2}{3}\times 0.975=0.65moles$ of $Fe_2O_3$

Mass of $Fe_2O_3=moles\times {\text {Molar mass}}=0.65moles\times 159.69g/mol=103.8g$

Thus 103.8 g of $Fe_2O_3$ will be produced.