If 100 g of iron reacts with excess oxygen,how much rust should be produced
Rust is a complex hydrated iron oxide. For purposes of this problem we'll call it Fe2O3.
..................4Fe + 3O2 ==> 2Fe2O3
mols Fe = grams/atomic mass = 100/55.85 = 1.8
Convert mols Fe to mols Fe2O3 this way.
1.8 mols Fe x (2 mols Fe2O3/4 mols Fe) = 0.9 mols Fe2O3
grams Fe2O3 = mols Fe2O3 x molar mass Fe2O3 = ?
To determine how much rust is produced when 100 g of iron reacts with excess oxygen, we first need to understand the balanced chemical equation for the reaction.
The balanced equation for the reaction between iron and oxygen is:
4 Fe + 3 O₂ → 2 Fe₂O₃
From the equation, we can see that 4 moles of iron (Fe) react with 3 moles of oxygen (O₂) to produce 2 moles of iron(III) oxide (Fe₂O₃). Now we need to calculate the number of moles of iron present in 100 g of iron.
The molar mass of iron (Fe) is approximately 55.845 g/mol. Therefore, the number of moles of iron can be calculated using the formula:
Number of moles = Mass (in grams) / Molar mass
Number of moles of iron = 100 g / 55.845 g/mol
Number of moles of iron ≈ 1.79 mol
Since the reaction between iron and oxygen occurs in a 4:3 ratio, we can use stoichiometry to determine the moles of iron(III) oxide (Fe₂O₃) produced.
According to the balanced equation, for every 4 moles of iron (Fe), 2 moles of iron(III) oxide (Fe₂O₃) are produced. Therefore, when 1.79 moles of iron react, the amount of iron(III) oxide produced can be calculated using the formula:
Mo
To determine how much rust would be produced when 100 g of iron reacts with excess oxygen, we need to consider the stoichiometry of the reaction.
The chemical equation for the reaction between iron and oxygen to form rust is:
4 Fe + 3 O2 -> 2 Fe2O3
From the balanced equation, we can see that for every 4 moles of iron (Fe) that react, we will get 2 moles of rust (Fe2O3) produced.
To find the molar mass of iron (Fe), we can refer to the periodic table, which tells us that the atomic mass of iron is 55.845 g/mol.
Now, let's calculate the amount of moles of iron in 100 g:
moles of Fe = mass of Fe / molar mass of Fe
= 100 g / 55.845 g/mol
≈ 1.79 mol
Since the stoichiometry of the reaction tells us that for every 4 moles of iron, we get 2 moles of rust, we can use this ratio to determine the moles of rust produced:
moles of Fe2O3 = (moles of Fe / 4) * (2 moles of Fe2O3 / 1 mole of Fe)
= (1.79 mol / 4) * 2
≈ 0.895 mol
Finally, to determine the mass of rust produced, we can use the molar mass of Fe2O3, which is 159.69 g/mol:
mass of Fe2O3 = moles of Fe2O3 * molar mass of Fe2O3
≈ 0.895 mol * 159.69 g/mol
≈ 142.87 g
Therefore, when 100 g of iron reacts with excess oxygen, approximately 142.87 g of rust should be produced.