How many moles of oxygen reacts with 2.4 moleof iron in the rusting reaction?
Rust is a somewhat complicated compound but if we call it simply Fe2O3, then 4Fe + 3O2 ==> 2Fe2O3
2.4 mols Fe x (3 mols O2/4 mols Fe) = 2.4 x 3/4 = ?
Well, it seems like the moles are breaking out in an absolute frenzy! In the rusting reaction, each mole of iron reacts with one mole of oxygen to form rust, otherwise known as iron(III) oxide. So, if you have 2.4 moles of iron, you'll need 2.4 moles of oxygen to dance with it. Just make sure they don't get too rusty on the dance floor!
In order to determine the number of moles of oxygen that react with 2.4 moles of iron in the rusting reaction, we need to balance the chemical equation first.
The rusting reaction can be represented by the following balanced equation:
4 Fe + 3 O2 -> 2 Fe2O3
From the balanced equation, we can see that 4 moles of iron react with 3 moles of oxygen to form 2 moles of iron(III) oxide (Fe2O3).
To calculate the moles of oxygen that react with 2.4 moles of iron, we can set up a proportion:
4 moles of iron / 3 moles of oxygen = 2.4 moles of iron / x moles of oxygen
Cross-multiplying, we get:
4 moles of iron * x moles of oxygen = 3 moles of oxygen * 2.4 moles of iron
Simplifying, we have:
4x = 7.2
Dividing both sides by 4, we find:
x = 1.8
Therefore, 1.8 moles of oxygen react with 2.4 moles of iron in the rusting reaction.
To determine how many moles of oxygen react with 2.4 moles of iron in the rusting reaction, we first need to know the balanced chemical equation for the reaction.
The balanced chemical equation for the rusting of iron can be represented as:
4 Fe + 3 O2 → 2 Fe2O3
From the balanced equation, we can see that each mole of iron (Fe) reacts with 3 moles of oxygen (O2) to produce 2 moles of iron(III) oxide (Fe2O3).
Using the stoichiometry of the balanced equation, we can set up a proportion to find the number of moles of oxygen reacting with 2.4 moles of iron:
(3 moles O2) / (4 moles Fe) = (x moles O2) / (2.4 moles Fe)
Cross-multiplying and solving for x, we get:
(3 moles O2) * (2.4 moles Fe) = (4 moles Fe) * (x moles O2)
7.2 moles O2 = 4x
Dividing both sides by 4:
7.2 / 4 = x
x = 1.8 moles O2
Therefore, in the rusting reaction, 1.8 moles of oxygen react with 2.4 moles of iron.