Problem 64 - Chemistry The Science in Context

Question

Chromium metal can be produced from the high-temperature reaction of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) [chromium(III) oxide] with silicon or aluminum by each of the following reactions:$$\begin{aligned} \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(\ell) & \rightarrow 2 \mathrm{Cr}(\ell)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) \\\2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+3 \mathrm{Si}(\ell) & \rightarrow 4 \mathrm{Cr}(\ell)+3 \mathrm{SiO}_{2}(s)\end{aligned}$$.a. Calculate the number of grams of aluminum required to prepare \(400.0 \mathrm{g}\) of chromium metal by the first reaction. b. Calculate the number of grams of silicon required to prepare \(400.0 \mathrm{g}\) of chromium metal by the second reaction.

Step-by-Step Solution

Answer

Question: Calculate the mass of aluminum and silicon required to produce 400.0 g of chromium metal using the given reactions. Answer: The mass of aluminum required to prepare 400.0 g of chromium metal by the first reaction is 207.5 g, and the mass of silicon required to prepare 400.0 g of chromium metal by the second reaction is 162.0 g.

Step 1: Write the balanced chemical equations

The balanced chemical equations are already provided: 1. \(Cr_{2}O_{3}(s) + 2Al(\ell) \rightarrow 2Cr(\ell) + Al_{2}O_{3}(s)\) 2. \(2Cr_{2}O_{3}(s) + 3Si(\ell) \rightarrow 4Cr(\ell) + 3SiO_{2}(s)\)

Step 2: Calculate the moles of chromium metal

To find the moles of chromium produced, we'll use the molar mass of chromium. The molar mass of chromium is 51.996 g/mol. Given mass of chromium = 400.0 g \(\text{Moles of chromium} = \frac{\text{Given mass of chromium}}{\text{Molar mass of chromium}} = \frac{400.0\,\text{g}}{51.996\,\text{g/mol}} = 7.692\,\text{mol} \)

Step 3: Calculate the moles of aluminum and silicon required

Now, we'll use the stoichiometry from the balanced chemical equations to calculate the moles of aluminum and silicon required for each reaction respectively: a. For the first reaction: \(Cr_{2}O_{3}(s) + 2Al(\ell) \rightarrow 2Cr(\ell) + Al_{2}O_{3}(s)\) From the stoichiometry, 2 moles of Al are required to produce 2 moles of Cr. \(\text{Moles of aluminum} = \text{Moles of chromium} = 7.692\,\text{mol} \) b. For the second reaction: \(2Cr_{2}O_{3}(s) + 3Si(\ell) \rightarrow 4Cr(\ell) + 3SiO_{2}(s)\) From the stoichiometry, 3 moles of Si are required to produce 4 moles of Cr. \(\text{Moles of silicon} = \frac{3}{4} × \text{Moles of chromium} = \frac{3}{4} \times 7.692\,\text{mol} = 5.769\,\text{mol}\)

Step 4: Calculate the mass of aluminum and silicon required

We'll use the molar masses of aluminum and silicon to calculate the mass required for each reaction: a. For the first reaction: Molar mass of aluminum = 26.98 g/mol \(\text{Mass of aluminum} = \text{Moles of aluminum} \times \text{Molar mass of aluminum} = 7.692\,\text{mol} \times 26.98\,\text{g/mol} = 207.5\,\text{g}\) b. For the second reaction: Molar mass of silicon = 28.09 g/mol \(\text{Mass of silicon} = \text{Moles of silicon} \times \text{Molar mass of silicon} = 5.769\,\text{mol} \times 28.09\,\text{g/mol} = 162.0\,\text{g}\)

Step 5: Conclusion

The mass of aluminum required to prepare 400.0 g of chromium metal by the first reaction is 207.5 g, and the mass of silicon required to prepare 400.0 g of chromium metal by the second reaction is 162.0 g.

Key Concepts

Understanding Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into different substances, called products. The transformations occur due to the breaking and forming of chemical bonds. To understand chemical reactions, one must be familiar with chemical equations that represent these reactions. A balanced chemical equation has the same number of atoms of each element on both sides of the reaction arrow, indicating that mass is conserved during the reaction.

For example, in the reaction where chromium metal is produced by reacting chromium(III) oxide with aluminum, the balanced chemical equation is: \(Cr_{2}O_{3}(s) + 2Al(l) \rightarrow 2Cr(l) + Al_{2}O_{3}(s)\). Each side of the equation has 2 chromium atoms, 3 oxygen atoms, and 2 aluminum atoms, showing it is balanced and obeys the Law of Conservation of Mass. This equation serves as a map for stoichiometry, allowing us to calculate the reactants needed to form a certain amount of product.

Molar Mass Calculation

The molar mass of a substance is defined as the mass of one mole (6.022 x 10²³ entities) of that substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule. For example, the molar mass of chromium (Cr) is 51.996 g/mol.

Understanding how to calculate molar mass is critical because it allows us to convert from the mass of a substance to the number of moles, serving as a bridge between the macroscopic world we can measure and the microscopic world of molecules and atoms. In essence, knowing the molar mass helps us to quantify substances in a chemical reaction accurately.

Mole-to-Mass Conversion

Mole-to-mass conversion is a pivotal concept in stoichiometry. It involves using the molar mass of a substance as a conversion factor to translate between the number of moles and the mass in grams. The process is straightforward once you have determined the molar mass of the substance in question.

To carry out a mole-to-mass conversion, you multiply the number of moles by the molar mass of the substance. For instance, in the reaction to produce chromium metal, the number of moles of aluminum required to react with chromium(III) oxide can be converted to grams by using the molar mass of aluminum: \(7.692\,\text{mol} \times 26.98\,\text{g/mol} = 207.5\,\text{g}\). This conversion is fundamental when preparing reagents for reactions or analyzing the results of a chemical experiment.