Question

Before arc welding was developed, a displacement reaction involving aluminum and iron(III) oxide was commonly used to produce molten iron (the thermite process). This reaction was used, for example, to connect sections of iron railroad track. Calculate the mass of molten iron produced when 2.28 kg of aluminum reacts with 17.2 mol of iron(III) oxide.

Answer 1

Answer: The mass of molten iron formed will be 1.92 kg

Step-by-step explanation:

We are given:

Moles of iron (III) oxide = 17.2 moles

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 aluminium = 2.28 kg = 2280 g (Conversion factor: 1 kg = 1000 g)

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

`\text{Moles of aluminium}=(2280g)/(27g/mol)=84.44mol`

The chemical equation for the reaction of aluminium and iron (III) oxide follows:

`2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe`

By Stoichiometry of the reaction:

1 mole of iron (III) oxide reacts with 2 moles of aluminium

So, 17.2 moles of iron (III) oxide will react with =
`(2)/(1)* 17.2=34.4mol` of aluminium

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

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

By Stoichiometry of the reaction:

1 mole of iron (III) oxide produces 2 moles of iron

So, 17.2 moles of iron (III) oxide will produce =
`(2)/(1)* 17.2=34.4mol` of iron

Now, calculating the mass of iron by using equation 1, we get:

Molar mass of iron = 55.8 g/mol

Moles of iron = 34.4 moles

Putting values in equation 1, we get:

`34.4mol=\frac{\text{Mass of iron}}{55.8g/mol}\\\\\text{Mass of iron}=(34.4mol* 55.8g/mol)=1919.5g=1.92kg`

Hence, the mass of molten iron formed will be 1.92 kg