Question

Iron rusts when exposed to oxygen. Iron(II) is produced in the reaction. How many moles of oxygen oxide will react with 2.4 mol of iron in the rusting reaction?

Answer 1

$\mathsf{\Large{\underline{Stoichiometry}}}$ :

$\mathsf{\large{\underline{Solution}}}$ :

Reaction involved :

$\boxed{\mathsf{Fe \:+\:{\dfrac{1}{2}}O2\:{\rightarrow{\:Fe( II ) O}}}}$

By using Unitary method,

Given, Moles of iron reacting ( Fe ) = 2.4 mol

Now,

2.4 mol of Iron reacts with $\mathsf{\dfrac{1}{2}}$ oxygen dioxide to produce 2.4 mol of FeO.

So, Number of moles of Oxygen dioxide produced = $\mathsf{\dfrac{2.4}{2}}$

↪️ Number of moles of Oxygen dioxide produced = 1.2 moles