Question

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?

Answer 1

136g of iron oxide will be produced

Answer 2

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

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For iron:

Given mass of iron = 112 g

Molar mass of iron = 55.84 g/mol

Putting values in equation 1, we get:

`\text{Moles of iron}=(112g)/(55.84g/mol)=2mol`

• For oxygen gas:

Given mass of oxygen gas = 24 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(24g)/(32g/mol)=0.75mol`

The given chemical equation follows:

`4Fe+3O_2\rightarrow 2Fe_2O_3`

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 =
`(3)/(4)* 0.75=0.5625mol` 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 =
`(2)/(4)* 0.75=0.375mol` 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:

`0.375mol=\frac{\text{Mass of iron (III) oxide}}{159.7g/mol}\\\\\text{Mass of iron (III) oxide}=(0.375mol* 159.7g/mol)=59.9g`

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