2. when iron reacts with oxygen, it forms iron oxide, or rust.
4 fe + 3 o2 2 fe2o3

if 112 g of iron (fe) combines with 24 g of oxygen (o2), how much iron oxide is formed?

The mass of iron (III) oxide produced is 59.9 grams.

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of iron = 112 g

Molar mass of iron = 55.84 g/mol

Putting values in equation 1, we get:

Given mass of oxygen gas = 24 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

The given chemical equation follows:

By Stoichiometry of the reaction:

4 moles of iron reacts with 3 moles of oxygen gas

So, 0.75 moles of iron will react with = of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, iron is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of iron produces 2 moles of iron (III) oxide

So, 0.75 moles of iron will produce = of iron (III) oxide

Now, calculating the mass of iron (III) oxide by using equation 1:

Molar mass of iron (III) oxide = 159.7 g/mol

Moles of iron (III) oxide = 0.375 moles

Putting values in equation 1, we get:

Hence, the mass of iron (III) oxide produced is 59.9 grams.