How many grams of iron must be oxidized to produce 5.0g of rust? (Fe2O3)
I have no idea how to solve this!
If your reaction is
4Fe + 3O2 -> 2Fe2O3
then it takes 2 moles of Fe to produce 1 mole of rust.
5.0g of rust is 0.0313 moles
So, how many grams of Fe in 0.626 moles?
would it be 10? since it takes double the amount of Fe to make rust. 5.0g of rust is .0313 moles, you would multiply it by 2?
1. Steve moved the decimal point; that should be 0.0626 mols.
2. John, that answer is not right.
g = mols x molar mass = 0.0626 x 55.85 = ?
To solve this problem, we need to first determine the stoichiometry of the reaction between iron (Fe) and oxygen (O2) to form rust (Fe2O3).
The balanced equation for this reaction is:
4 Fe + 3 O2 -> 2 Fe2O3
According to the stoichiometry, 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3.
Now, we can calculate the molar mass of Fe2O3:
Fe: 55.845 g/mol
O: 15.999 g/mol (There are 3 oxygen atoms in Fe2O3, so we multiply the molar mass by 3)
Molar Mass of Fe2O3 = (2 × 55.845 g/mol) + (3 × 15.999 g/mol) = 159.688 g/mol
Next, we need to use the molar mass of Fe2O3 to convert the given mass of rust (5.0g) into moles. We can use the following formula:
moles = mass / molar mass
Moles of Fe2O3 = 5.0 g / 159.688 g/mol ≈ 0.031 moles
Finally, to find the number of grams of iron that would be oxidized, we use the stoichiometry of the balanced equation. From the equation, we know that 2 moles of Fe2O3 are formed from 4 moles of Fe. Therefore, we can write the following proportion:
(2 moles Fe2O3) / (4 moles Fe) = (0.031 moles Fe2O3) / (x moles Fe)
Cross-multiplying and solving for x, we get:
x = (4 moles Fe * 0.031 moles Fe2O3) / 2 moles Fe2O3 ≈ 0.062 moles Fe
Finally, we can convert moles of Fe to grams using the molar mass of Fe:
mass of Fe = moles × molar mass
mass of Fe = 0.062 moles × 55.845 g/mol ≈ 3.49 g
Therefore, approximately 3.49 grams of iron must be oxidized to produce 5.0 grams of rust (Fe2O3).