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)#?

1 Answer
Mar 28, 2018

Answer:

#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#