When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 7.2 mol of iron in the rusting reaction? 4Fe(s)+3O2=2Fe2O3(s)
To determine the number of moles of oxygen that react with 7.2 moles of iron in the rusting reaction, we will use the stoichiometric coefficients from the balanced chemical equation.
In the balanced equation: 4Fe(s) + 3O2 (g) -> 2Fe2O3(s)
The ratio of moles of iron to moles of oxygen is 4:3. This means that for every 4 moles of iron that react, 3 moles of oxygen are required.
Given that we have 7.2 moles of iron, we can calculate the moles of oxygen using the ratio above:
7.2 moles iron * (3 moles oxygen / 4 moles iron) = 5.4 moles of oxygen
Therefore, 5.4 moles of oxygen react with 7.2 moles of iron in the rusting reaction.
To determine the number of moles of oxygen that react with 7.2 mol of iron, we can use the balanced chemical equation provided. The coefficient in front of oxygen (O2) in the equation tells us the stoichiometric ratio between iron and oxygen.
The balanced equation is: 4Fe(s) + 3O2(g) → 2Fe2O3(s)
From the equation, we can see that for every 4 moles of iron (Fe) that react, 3 moles of oxygen (O2) are required. So, the stoichiometric ratio of iron to oxygen is 4:3.
Now we can calculate the moles of oxygen. We know that we have 7.2 moles of iron (Fe). Using the stoichiometric ratio, we can set up a simple proportion:
(3 mol O2 / 4 mol Fe) = (x mol O2 / 7.2 mol Fe)
Simplifying the equation, we get:
3 * 7.2 = 4 * x
21.6 = 4x
Now, solve for x:
x = 21.6 / 4
x ≈ 5.4
Therefore, approximately 5.4 moles of oxygen react with 7.2 moles of iron in the rusting reaction.
Use the coefficients in the balanced equation to convert anything to anything.
7.2 mol Fe x (3 mols O2/4 mols Fe) = 7.2*3/4 = ?