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 15.0g of Iron, we need 21.4g of iron(III) oxide and 7.22g of aluminum. The maximum mass of aluminum oxide that could be produced is 13.7g.

Step-by-step solution

Write down the balanced chemical equation

: Given the balanced chemical equation: \[Fe_2 O_3 (s) + 2 Al (s) \rightarrow 2 Fe (l) + Al_2 O_3 (s)\]

Calculate the moles of Iron needed

: First, let's convert the mass (15.0g) of iron to moles. To do this, we will use the molar mass of iron: \(1 mol \, of \, Fe = 55.85 g\). Moles of Iron = mass of Iron / molar mass of Fe \[moles \, of \, Fe = \frac{15.0g}{55.85 g/mol} = 0.268mol\]

Determine the moles of Iron(III) oxide and Aluminum required

: Using the stoichiometric ratios from the balanced chemical equation, we can calculate the moles of Fe2O3 and Al needed. Moles of Fe2O3 = Moles of Fe x (1 mol of Fe2O3 / 2 mol of Fe) \[ moles \, of \, Fe_2 O_3 = 0.268mol \times \frac{1}{2} = 0.134mol\] Moles of Al = Moles of Fe x (2 mol of Al / 2 mol of Fe) \[moles \, of \, Al = 0.268mol \times \frac{2}{2} = 0.268mol\]

Convert moles to mass

: To find the masses of iron(III) oxide and aluminum, we can use their molar masses: \(1 mol \, of \, Fe_2 O_3 = 159.69 g\) and \(1 mol \, of \, Al = 26.98 g\). Mass of Fe2O3 = moles of Fe2O3 x molar mass of Fe2O3 \[mass\, of\, Fe_2 O_3 = 0.134mol \times 159.69 g/mol = 21.4 g\] Mass of Al = moles of Al x molar mass of Al \[mass\, of\, Al = 0.268mol \times 26.98 g/mol = 7.22 g\]

Calculate the maximum mass of Aluminum oxide

: We can use the stoichiometric ratio of Al2O3 to Fe to compute the maximum mass of Al2O3 produced. Moles of Al2O3 = moles of Fe x (1 mol of Al2O3 / 2 mol of Fe) \[moles\, of\, Al_2 O_3 = 0.268mol \times \frac{1}{2} = 0.134mol\] Now, we will convert the moles of Al2O3 to mass using its molar mass: \(1 mol \, of \, Al_2 O_3 = 101.96 g\). Mass of Al2O3 = moles of Al2O3 x molar mass of Al2O3 \[mass\, of\, Al_2 O_3 = 0.134 mol \times 101.96 g/mol = 13.7 g\] To produce 15.0g of Iron, we need 21.4g of iron(III) oxide and 7.22g of aluminum. The maximum mass of aluminum oxide that could be produced is 13.7g.

Key concepts

Stoichiometric Calculations

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Specifically, it involves calculations to determine the amounts of substances that are consumed and produced in a reaction. For instance, the thermite reaction uses iron(III) oxide and aluminum to produce iron and aluminum oxide.

When solving stoichiometric problems, the first step is to ensure you have a balanced chemical equation. This equation tells you the ratio in which reactants combine and products form, often referred to as the stoichiometric coefficients. In our thermite example, the balanced equation is: \[Fe_2 O_3 (s) + 2 Al (s) \rightarrow 2 Fe (l) + Al_2 O_3 (s)\].

These coefficients are used to convert between moles of one substance to moles of another. By understanding the stoichiometry of the reaction, we can answer questions like how much aluminum is needed to react with a certain mass of iron(III) oxide, or how much aluminum oxide is produced when a set amount of iron is formed.

Chemical Equation Balancing

Balancing chemical equations is essential for performing stoichiometric calculations. An equation must reflect the conservation of mass, which means the number of atoms for each element in the reactants must equal the number in the products. To achieve this balance, we adjust coefficients in front of the chemical formulas without changing the formulas themselves.

Looking at our thermite equation again, the balanced form is: \[Fe_2 O_3 (s) + 2 Al (s) \rightarrow 2 Fe (l) + Al_2 O_3 (s)\].

This shows that one molecule of iron(III) oxide reacts with two atoms of aluminum to produce two atoms of iron and one molecule of aluminum oxide. Balancing is the foundation for stoichiometry, as it ensures the mole ratios needed for calculations represent the actual consumption of reactants and formation of products.

Molar Mass

The molar mass of a substance is the weight of one mole of that substance, typically expressed in grams per mole (\(g/mol\)). It is a crucial value for converting between the weight of a substance and the amount in moles, which is the basis for all stoichiometric calculations.

For example, iron (Fe) has a molar mass of 55.85\(g/mol\). This means that one mole, or 6.022 x 10²³ atoms, of iron weighs 55.85 grams. When given a mass of iron, like the 15.0 grams in our thermite problem, we can use its molar mass to find the number of moles. These molar mass values of reactants and products are essential for step-by-step stoichiometric calculations.

Mole-to-Mass Conversion

When working with stoichiometric problems, converting between moles and mass is a common task that relies on the molar mass of the substances involved. After determining the number of moles from a balanced equation, we can calculate the corresponding mass using molar mass.

In the thermite reaction scenario, we first converted the desired mass of iron into moles. Then, using the stoichiometric coefficients and molar masses of iron(III) oxide (159.69\(g/mol\)) and aluminum (26.98\(g/mol\)), we converted moles back into mass to find out how much of each reactant is needed. This process is fundamental for predicting the amounts of materials required or produced in a given reaction.