iron reacts with oxygen to form iron oxide. if 100g of iron reacts with excess oxygen, how much rust should be produced?

1 answer

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.