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?

Question

asked Dec 27, 2018 125k views

2 Answers

OlliM · Answer 1 · 2019-01-01T06:40:38+0000

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)

Given mass of iron = 112 g

Molar mass of iron = 55.84 g/mol

Putting values in equation 1, we get:

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

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.