Question

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ What masses of iron(III) oxide and aluminum must be used to produce 15.0 \(\mathrm{g}\) iron? What is the maximum mass of aluminum oxide that could be produced?

Short answer

To produce 15.0 g of iron, 21.45 g of iron(III) oxide and 14.49 g of aluminum must be used. The maximum mass of aluminum oxide that could be produced is 54.75 g.

Step-by-step solution

Determine the moles of iron required

First, let's convert the given mass of iron (15.0 g) into moles using its molar mass. The molar mass of iron (Fe) is 55.85 g/mol. Moles of iron = \(\frac{15.0\,\text{g}}{55.85\,\text{g/mol}}\) = 0.2685 mol

Determine the moles of iron(III) oxide and aluminum required

Now, let's determine the moles of iron(III) oxide (Fe2O3) and aluminum (Al) needed for the reaction, using the stoichiometry of the balanced chemical equation. Moles of Fe2O3 = \(\frac{1\,\text{mol of Fe2O3}}{2\,\text{mol of Fe}} × 0.2685\,\text{mol of Fe}\) = 0.1343 mol of Fe2O3 Moles of Al = \(\frac{2\,\text{mol of Al}}{2\,\text{mol of Fe}} × 0.2685\,\text{mol of Fe}\) = 0.537 mol of Al

Calculate the masses of iron(III) oxide and aluminum required

To calculate the masses of the reactants, we will multiply the moles obtained in Step 2 by their corresponding molar masses. Mass of Fe2O3 = 0.1343 mol of Fe2O3 × \(159.69\,\text{g/mol}\) = 21.45 g of Fe2O3 Mass of Al = 0.537 mol of Al × 26.98 g/mol = 14.49 g of Al

Calculate the maximum mass of aluminum oxide produced

Lastly, we will find the maximum mass of aluminum oxide (Al2O3) that can be produced by multiplying the moles of aluminum used (0.537 mol) by the molar mass of Al2O3. Mass of Al2O3 = 0.537 mol of Al × \(101.96\,\text{g/mol}\) = 54.75 g of Al2O3 #Conclusion# To produce 15.0 g of iron, 21.45 g of iron(III) oxide and 14.49 g of aluminum must be used. The maximum mass of aluminum oxide that could be produced is 54.75 g.

Key concepts

Stoichiometry

Stoichiometry is the heart of balancing and calculating quantities in chemical reactions. It helps determine how much of each reactant is needed to form the products in a balanced chemical equation. In the case of the thermite reaction, stoichiometry tells us how the quantities of iron(III) oxide and aluminum relate to the desired amount of iron being produced. By understanding the molar ratios given in the balanced equation, we can determine the proportions of reactants needed. For example, in this reaction, 1 mole of iron(III) oxide reacts with 2 moles of aluminum to produce 2 moles of iron. This crucial relationship allows us to convert moles needed into actual masses.

Molar Mass

Molar mass is a foundational concept that connects the mass of a substance to the number of moles. It is essential for converting between grams and moles, as demonstrated in step-by-step chemical problem-solving. Molar mass is expressed in grams per mole (g/mol) and is critical for calculating how much of each element or compound is involved in a reaction. For example, in our exercise, the molar mass of iron (Fe) is 55.85 g/mol. This information is used to convert the mass of iron needed for the reaction (15.0 g) into moles. Knowing the molar masses of iron(III) oxide (159.69 g/mol), aluminum (26.98 g/mol), and aluminum oxide (101.96 g/mol) allows us to find out how much of each reactant is required, and the expected product yield.

Chemical Equation

A chemical equation symbolizes a chemical reaction clearly and concisely. It shows how reactants transform into products, indicating the proportions in which the substances unite or break apart. In the thermite reaction equation \( \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + 2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(l) + \mathrm{Al}_{2} \mathrm{O}_{3}(s) \), the arrow signifies the direction of the reaction from reactants to products. It is essential for the equation to be balanced, meaning the same number of atoms of each element on both sides of the equation, to faithfully represent the conservation of mass. With this balance, we can apply stoichiometry accurately.

Reactants

Reactants are the starting materials in a chemical reaction, found on the left side of a chemical equation. They undergo chemical changes during the reaction to form new substances. In the thermite reaction, the reactants are iron(III) oxide \( (\mathrm{Fe}_{2}\mathrm{O}_{3} ) \) and aluminum \( (\mathrm{Al}) \). Each plays a pivotal role in the progression of the reaction. Understanding the needs of the reactants—like their molar amounts and mass—is vital for predicting how much of each is necessary to yield a desired amount of product. For instance, to produce 15.0 g of iron, specific masses of \( \mathrm{Fe}_{2}\mathrm{O}_{3} \) and \( \mathrm{Al} \) need to be accurately calculated and used.

Products

Products are the outcome of a chemical reaction, appearing on the right side of a chemical equation. They are the new substances formed when reactants undergo chemical change. In the provided thermite reaction, the products are molten iron \( (\mathrm{Fe}) \) and aluminum oxide \( (\mathrm{Al}_{2}\mathrm{O}_{3}) \). Knowing the desired amount of product—in this case, 15.0 g of iron—helps determine how much of the reactants need to be used. Additionally, result predictions for by-products like aluminum oxide involve stoichiometric calculations using molar masses, enabling comprehensive understanding and planning of reaction outputs.