When iron rusts in air, iron(III) oxide is produced. How many moles of oxygen react with 2.4 mol of iron in the rusting reaction?

1.8 moles

Meg--All of these stoichiometry problems are worked alike. I gave you the equation for iron and oxygen in the example I worked. The problem gives you mols so you don't need step 2 to convert grams to moles; that has been done for you.

The balanced chemical equation for the rusting of iron can be written as:

4 Fe + 3 O2 → 2 Fe2O3

From this equation, we can see that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron(III) oxide.

Given that 2.4 moles of iron are involved in the reaction, we can use the mole ratio from the balanced equation to determine the number of moles of oxygen required.

4 moles Fe = 3 moles O2
2.4 moles Fe = (3/4) x 2.4 moles O2

Calculating this gives:

2.4 moles Fe = 1.8 moles O2

Therefore, 1.8 moles of oxygen would react with 2.4 moles of iron in the rusting reaction.

To determine how many moles of oxygen react with 2.4 mol of iron in the rusting reaction, we need to look at the balanced chemical equation for the reaction between iron and oxygen to form iron(III) oxide.

The balanced equation for the reaction is:
4 Fe + 3 O2 → 2 Fe2O3

From the equation, it is clear that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron(III) oxide.

Now, we can use this ratio to calculate the number of moles of oxygen that would react with 2.4 mol of iron.

Given: Moles of iron = 2.4 mol

Using the stoichiometric ratio from the balanced equation, we have:
(3 mol O2 / 4 mol Fe) × 2.4 mol Fe = 1.8 mol O2

Therefore, 1.8 mol of oxygen would react with 2.4 mol of iron in the rusting reaction.