when iron rusts in air, iron(III) oxide is produced. how many moles react with 2.4 iron in the rusting reaction? 4Fe(s) + 3O2(g) => 2Fe2O3(s)

Answered below.

To determine the number of moles of iron(III) oxide produced in the rusting reaction, we need to use the stoichiometry of the balanced equation.

The balanced equation for the reaction is:

4Fe(s) + 3O2(g) -> 2Fe2O3(s)

From the equation, we can see that the ratio between iron and iron(III) oxide is 4:2. This means that for every 4 moles of iron, we get 2 moles of iron(III) oxide.

Given that we have 2.4 moles of iron, we can calculate the number of moles of iron(III) oxide produced using the ratio as follows:

2.4 moles of Fe * (2 moles of Fe2O3 / 4 moles of Fe) = 1.2 moles of Fe2O3

Therefore, 1.2 moles of iron(III) oxide will react with 2.4 moles of iron in the rusting reaction.

To find the number of moles of iron(III) oxide produced when 2.4 moles of iron reacts, we can use the balanced equation for the rusting reaction.

From the balanced equation: 4Fe(s) + 3O2(g) => 2Fe2O3(s)

We can see that 4 moles of iron react with 2 moles of iron(III) oxide.

To determine the number of moles that react with 2.4 moles of iron, we can set up a proportion:

4 moles of iron / 2 moles of iron(III) oxide = 2.4 moles of iron / x moles of iron(III) oxide

Cross-multiplying:

4 moles of iron * x moles of iron(III) oxide = 2.4 moles of iron * 2 moles of iron(III) oxide

Simplifying:

4x = 4.8

Dividing both sides by 4:

x = 1.2

Therefore, 1.2 moles of iron(III) oxide will react with 2.4 moles of iron in the rusting reaction.