Question

Before arc welding was developed, a displacement reaction involving aluminum and iron(III) oxide was commonly used to produce molten iron. Called the thermite process, this reaction was used, for example, to connect sections of iron rails for train tracks. Calculate the mass of molten iron produced when \(1.50 \mathrm{~kg}\) of aluminum reacts with \(25.0 \mathrm{~mol}\) of iron(III) oxide.

Short answer

2.79 kg of iron is produced.

Step-by-step solution

- Write the balanced chemical equation

First, write the balanced chemical equation for the thermite reaction:\[ 2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow 2\text{Fe} + \text{Al}_2\text{O}_3 \]

- Determine molar masses

Find the molar masses of aluminum (Al), iron(III) oxide (Fe_2O_3), and iron (Fe) using the periodic table:- Molar mass of Al = 26.98 g/mol- Molar mass of Fe_2O_3 = 159.69 g/mol- Molar mass of Fe = 55.85 g/mol

- Convert mass of aluminum to moles

Convert the given mass of aluminum (1.50 kg) to moles:\[ 1.50 \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}} \times \frac{1 \text{ mol Al}}{26.98 \text{ g}} = 55.58 \text{ mol Al} \]

- Find limiting reactant

Compare the mole ratio of aluminum to iron(III) oxide from the balanced equation with the provided amounts to identify the limiting reactant:- According to the balanced equation: 2 moles of Al react with 1 mole of Fe_2O_3- Given: 55.58 moles of Al and 25.0 moles of Fe_2O_3Since 25.0 moles of Fe_2O_3 would require 50.0 moles of Al (and we have more, 55.58 moles), Fe_2O_3 is the limiting reactant.

- Calculate moles of iron produced

Use the stoichiometry of the balanced equation to determine the moles of iron produced:- According to the balanced equation: 1 mole of Fe_2O_3 produces 2 moles of Fe- Given: 25.0 moles of Fe_2O_3- Moles of Fe produced: \[ 25.0 \text{ mol Fe}_2\text{O}_3 \times \frac{2 \text{ mol Fe}}{1 \text{ mol Fe}_2\text{O}_3} = 50.0 \text{ mol Fe} \]

- Convert moles of iron to mass

Convert the moles of iron produced to mass:\[ 50.0 \text{ mol Fe} \times \frac{55.85 \text{ g}}{1 \text{ mol Fe}} = 2792.5 \text{ g Fe}\]Convert grams to kilograms for the final answer: \[ 2792.5 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 2.79 \text{ kg Fe} \]

Key concepts

stoichiometry

Stoichiometry is a cornerstone concept in chemistry that involves calculating the quantities of reactants and products in a chemical reaction. It relies on the principle of the conservation of mass and the concept of moles, which represent quantities of particles. The key steps in stoichiometric calculations include:

• Writing a balanced chemical equation to understand the relationships between different substances.
• Using molar ratios from the balanced equation to convert between different reactants and products.

In the thermite process, the balanced equation is:\[ 2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow 2\text{Fe} + \text{Al}_2\text{O}_3 \]
This equation shows that two moles of aluminum react with one mole of iron(III) oxide to produce two moles of iron and one mole of aluminum oxide.

limiting reactants

The concept of limiting reactants is crucial to understand because it determines the maximum amount of product that can be formed in a chemical reaction. The limiting reactant is the substance that is completely consumed first, stopping the reaction from continuing.

To find the limiting reactant:

• Compare the mole ratio of the reactants provided to those in the balanced equation.
• Identify which reactant will be used up first based on these ratios.

From the exercise, given 55.58 moles of aluminum and 25.0 moles of iron(III) oxide:

• According to the balanced equation, 2 moles of aluminum react with 1 mole of iron(III) oxide.
• Therefore, 25.0 moles of iron(III) oxide would require 50.0 moles of aluminum.

As we have more aluminum (55.58 moles) than needed, the iron(III) oxide is the limiting reactant.

molar mass calculation

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Calculating the molar mass is fundamental for converting between the mass of a substance and the number of moles.

To find the molar mass, sum the atomic masses of all atoms in the molecular formula. For example:

• Aluminum (Al): 26.98 g/mol.
• Iron(III) oxide (Fe_2O_3): \[ 2(\text{Fe: 55.85 g/mol}) + 3 (\text{O: 16.00 g/mol}) = 159.69 \text{ g/mol} \]
• Iron (Fe): 55.85 g/mol.

In our exercise, you need to convert the mass of aluminum to moles:

• \[ 1.50 \text{ kg} \times 1000 \text{ g/kg} = 1500 \text{ g} \]
• \[ 1500 \text{ g} \times \frac{1 \text{ mol Al}}{26.98 \text{ g}} = 55.58 \text{ mol Al} \]

Understanding molar masses and practicing these calculations are crucial for mastering stoichiometry and solving chemical reaction problems confidently.