Question

When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 12.9 mol of iron in the rusting reaction? `4Fe(s) + 3O_2(g) -> 2Fe_2O_3(s)`?

Answer 1

`9.675 \ "mol"` of oxygen gas.

Explanation:

We have the balanced equation:

`4Fe(s)+3O_2(g)->2Fe_2O_3(s)`

And so, `4` moles of iron react with `3` moles of oxygen, and therefore, `12.9 \ "mol"` of iron would react with:

`12.9color(red)cancelcolor(black)("mol" \ Fe)*(3 \ "mol" \ O_2)/(4color(red)cancelcolor(black)("mol" \ Fe))=9.675 \ "mol" \ O_2`