Question

In a reaction, \(34.0 \mathrm{~g}\) of chromium(III) oxide reacts with \(12.1 \mathrm{~g}\) of aluminum to produce chromium and aluminum oxide. If \(23.3 \mathrm{~g}\) of chromium is produced, what mass of aluminum \(\mathrm{ox}\) ide is produced?

Short answer

In the reaction, 34.0 g of chromium(III) oxide and 12.1 g of aluminum produce 23.3 g of chromium. The limiting reactant is chromium(III) oxide. Using stoichiometry, we find that 0.335 moles of aluminum oxide are produced, which corresponds to a mass of 34.2 g.

Step-by-step solution

Write the chemical equation

First, we need to write the balanced chemical equation for the reaction. The reactants are chromium(III) oxide (Cr₂O₃) and aluminum (Al). The products are chromium (Cr) and aluminum oxide (Al₂O₃). The balanced chemical equation is: \[2 \mathrm~Cr_2O_3 + 3 \mathrm~Al \rightarrow 4 \mathrm~Cr + 3 \mathrm~Al_2O_3\]

Calculate moles for the given mass

Next, we need to convert the given mass of reactants (Cr₂O₃ and Al) and the mass of product (Cr) into moles. Use the molar masses of elements for the conversion: Molar mass of Cr = 51.996 g/mol Molar mass of O = 16.00 g/mol Molar mass of Al = 26.98 g/mol Molar mass of Cr₂O₃ = (2 × 51.996) + (3 × 16.00) = 151.992 g/mol Molar mass of Al₂O₃ = (2 × 26.98) + (3 × 16.00) = 101.96 g/mol Now, calculate the moles: moles of Cr₂O₃ = \( \frac{34.0 \ \mathrm g}{151.992 \ \mathrm g/mol} \) = 0.223 moles moles of Al = \( \frac{12.1 \ \mathrm g}{26.98 \ \mathrm g/mol} \) = 0.449 moles moles of Cr = \( \frac{23.3 \ \mathrm g}{51.996 \ \mathrm g/mol} \) = 0.448 moles

Find the limiting reactant

According to the balanced chemical equation, the stoichiometry of the reaction is: 2 moles of Cr₂O₃ : 3 moles of Al : 4 moles of Cr : 3 moles of Al₂O₃ Compare the mole ratio of the reactants: \( \frac{0.223 \ \mathrm{moles \ of \ Cr_2O_3}}{0.449 \ \mathrm{moles \ of \ Al}} \) = 0.496 Comparing with the stoichiometric ratio, \( \frac{2 \ \mathrm{moles \ of \ Cr_2O_3}}{3 \ \mathrm{moles \ of \ Al}} \) = 0.667 Since 0.496 < 0.667, the limiting reactant is chromium(III) oxide (Cr₂O₃).

Calculate the moles of aluminum oxide produced

According to the balanced chemical equation, 2 moles of Cr₂O₃ will produce 3 moles of Al₂O₃. So, calculate the moles of aluminum oxide produced using the limiting reactant: moles of Al₂O₃ = \( \frac{3 \ \mathrm{moles \ of \ Al_2O_3}}{2 \ \mathrm{moles \ of \ Cr_2O_3}} \) × 0.223 moles of Cr₂O₃ = 0.335 moles

Calculate the mass of aluminum oxide produced

Finally, convert the moles of aluminum oxide (Al₂O₃) produced into mass using the molar mass of Al₂O₃: mass of Al₂O₃ = 0.335 moles × 101.96 g/mol = 34.2 g Hence, 34.2 g of aluminum oxide (Al₂O₃) is produced in this reaction.

Key concepts

Understanding Chemical Reactions

When we talk about chemical reactions, we're discussing the process where substances, called reactants, transform into new substances, known as products. This transformation occurs as atoms are rearranged through the breaking and forming of chemical bonds. The key to a chemical reaction is the balanced chemical equation, which provides a recipe for what reactants are needed and what products will form, along with their respective ratios.

For instance, in the presented exercise, chromium(III) oxide reacts with aluminum to produce chromium and aluminum oxide, as shown by the equation:
\[2 \mathrm{Cr_2O_3} + 3 \mathrm{Al} \rightarrow 4 \mathrm{Cr} + 3 \mathrm{Al_2O_3}\].
Each number in front of a chemical formula represents the coefficient, which indicates how many moles of each substance are involved. This concept is crucial because it tells us the proportion in which reactants combine and the amount of products that will form.

Demystifying the Mole Concept

The mole concept is a fundamental pillar in chemistry, as it allows chemists to work with the submicroscopic world of atoms and molecules in a practical way. One mole is defined as Avogadro's number (\(6.022 \times 10^{23}\)) of entities, whether they're atoms, ions, or molecules. The molar mass, which is the mass of one mole of a substance in grams, bridges the gap between the microscopic and macroscopic world.

In the context of the exercise, we use the molar mass to convert given masses of chromium(III) oxide and aluminum into moles. This step is crucial because stoichiometric calculations in chemical reactions are based on moles, not on grams. For example, the molar mass of chromium(III) oxide is \(151.992 \mathrm{g/mol}\) and is used to calculate the amount in moles, starting from the provided mass. Understanding this concept is essential for successfully navigating through stoichiometry problems.

Unraveling the Limiting Reactant Concept

When performing a reaction with a given amount of reactants, one of them will be consumed first, halting the reaction and determining the maximum amount of product that can be formed; this is known as the limiting reactant. Identifying the limiting reactant is a crucial step in stoichiometry calculations, as it dictates the extent of the reaction.

In the given exercise, we must first find the ratio of moles of each reactant and compare it to the stoichiometric ratios from the balanced chemical equation. For instance, although we have more moles of aluminum, the stoichiometry of the reaction means that chromium(III) oxide is the limiting reactant. Once it's consumed, no additional chromium or aluminum oxide can be produced, even if there's aluminum left.

This concept is not only pivotal in theoretical exercises but also in practical applications such as yield optimization in industrial chemical synthesis, where understanding which reactant limits the process can lead to cost savings and efficiency improvements.