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)#?

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)#?

Owen Ambrose · Accepted Answer

#9.675 \ "mol"# of oxygen gas.

Here is the equation that is balanced: #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#

Natalie Anton · Answer

undefined In the given reaction, 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron (III) oxide. Therefore, to find out how many moles of oxygen react with 12.9 moles of iron, we can set up a proportion:

4 moles Fe : 3 moles O₂ = 12.9 moles Fe : x moles O₂

Cross-multiplying, we get:

4 * x = 3 * 12.9
4x = 38.7
x = 38.7 / 4
x = 9.675

So, approximately 9.675 moles of oxygen react with 12.9 moles of iron in the rusting reaction.

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) -&gt; 2Fe_2O_3(s)#?

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)#?