125k views
3 votes
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?

2 Answers

1 vote
136g of iron oxide will be produced
User HHHH
by
8.0k points
5 votes

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.

User OlliM
by
7.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.