iron reacts with oxygen to form iron oxide. if 100g of iron reacts with excess oxygen, how much rust should be produced?
The balanced chemical equation for the reaction between iron and oxygen is:
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 oxide.
To calculate the amount of iron oxide (rust) produced, we need to determine the number of moles of iron. We can use the molar mass of iron to convert grams to moles.
Molar mass of iron (Fe) = 55.845 g/mol
Number of moles of iron = 100g / 55.845 g/mol = 1.79 moles
From the balanced equation, we know that 4 moles of iron react with 2 moles of iron oxide. Therefore, the molar ratio between iron and iron oxide is 4:2 or 1:0.5.
Number of moles of iron oxide (rust) produced = 1.79 moles * 0.5 = 0.895 moles
To calculate the mass of iron oxide produced, we can use the molar mass of iron oxide.
Molar mass of iron oxide (Fe2O3) = 159.69 g/mol
Mass of iron oxide (rust) produced = 0.895 moles * 159.69 g/mol = 143.07 g
Therefore, 100g of iron should produce 143.07g of rust.
To determine the amount of rust produced when 100g of iron reacts with excess oxygen, we need to calculate the stoichiometry of the reaction.
The balanced chemical 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 react with 3 moles of oxygen to produce 2 moles of iron oxide (rust).
To calculate the amount of rust produced, we need to convert the mass of iron to moles using the molar mass of iron, which is 55.85 g/mol.
Number of moles of iron = mass of iron / molar mass of iron
= 100g / 55.85g/mol
≈ 1.79 moles
Based on the stoichiometry of the reaction, we can determine how many moles of rust will be produced:
From the balanced equation, there is a 4:2 ratio of iron to rust, so the number of moles of iron oxide (rust) produced will be half of the number of moles of iron.
Number of moles of rust = 1.79 moles / 2
≈ 0.895 moles
Finally, we can calculate the mass of rust produced by multiplying the number of moles by the molar mass of iron oxide (rust), which is 159.69 g/mol.
Mass of rust produced = number of moles of rust x molar mass of rust
≈ 0.895 moles x 159.69 g/mol
≈ 142.8 g
Therefore, approximately 142.8 grams of rust should be produced when 100 grams of iron reacts with excess oxygen.