Problem 36 - Chemistry: An Atoms-Focused Approach

Question

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

Step-by-Step Solution

Answer

Question: Calculate the mass of aluminum and silicon required to prepare 400.0 grams of chromium metal by their respective reactions with chromium(III) oxide. Answer: The mass of aluminum required to prepare 400.0 grams of chromium metal is 207.5 grams, and the mass of silicon required is 162.0 grams.

Step 1: Write down the balanced chemical equation for the first reaction

The balanced equation relating chromium(III) oxide, aluminum, and chromium in this reaction is given as: $$\mathrm{Cr}_{2} \mathrm{O}_{3}(s) + 2 \mathrm{Al}(\ell) \to 2 \mathrm{Cr(\ell)} + \mathrm{Al}_{2} \mathrm{O}_{3}(s)$$

Step 2: Calculate the molar masses of chromium and aluminum

Convert the masses of chromium and aluminum into moles using their respective molar masses: - Molar mass of chromium (Cr) = 51.996 g/mol - Molar mass of aluminum (Al) = 26.982 g/mol

Step 3: Determine the moles of chromium and aluminum needed

We are given the target mass of chromium (400.0 g) and need to calculate the mass of aluminum required. Convert the mass of chromium to moles: $$\mathrm{Moles \ of \ Cr} = \frac{400.0 \ g \ \mathrm{Cr}}{51.996 \ g/mol \ \mathrm{Cr}} = 7.692 \ moles \ \mathrm{Cr}$$ Now, use the stoichiometry of the balanced equation (2 moles Al to 2 moles Cr) to find the number of moles of aluminum required: $$\mathrm{Moles \ of \ Al} = \frac{7.692 \ moles \ \mathrm{Cr} \times 2 \ moles \ \mathrm{Al}}{2 \ moles \ \mathrm{Cr}} = 7.692 \ moles \ \mathrm{Al}$$

Step 4: Calculate the mass of aluminum required

Now that we know the number of moles of aluminum needed, we can convert this to mass: $$\mathrm{Mass \ of \ Al} = 7.692 \ moles \ \mathrm{Al} \times 26.982 \ g/mol \ \mathrm{Al} = 207.5 \ g \ \mathrm{Al}$$ Thus, 207.5 grams of aluminum is needed to produce 400.0 grams of chromium in the first reaction. #b. Calculate the mass of silicon required to prepare 400.0 grams of chromium metal by the second reaction.#

Step 1: Write down the balanced chemical equation for the second reaction

The balanced equation relating chromium(III) oxide, silicon, and chromium in this reaction is given as: $$2 \ \mathrm{Cr}_{2} \mathrm{O}_{3}(s) + 3 \ \mathrm{Si}(\ell) \to 4 \ \mathrm{Cr}(\ell) + 3 \ \mathrm{SiO}_{2}(s)$$

Step 2: Calculate the molar mass of silicon

Convert the mass of silicon into moles using its molar mass: - Molar mass of silicon (Si) = 28.085 g/mol

Step 3: Determine the moles of chromium and silicon needed

We have already calculated the moles of chromium needed. Now, use the stoichiometry of the balanced equation (3 moles Si to 4 moles Cr) to find the number of moles of silicon required: $$\mathrm{Moles \ of \ Si} = \frac{7.692 \ moles \ \mathrm{Cr} \times 3 \ moles \ \mathrm{Si}}{4 \ moles \ \mathrm{Cr}} = 5.769 \ moles \ \mathrm{Si}$$

Step 4: Calculate the mass of silicon required

Now that we know the number of moles of silicon needed, convert this to mass: $$\mathrm{Mass \ of \ Si} = 5.769 \ moles \ \mathrm{Si} \times 28.085 \ g/mol \ \mathrm{Si} = 162.0 \ g \ \mathrm{Si}$$ Thus, 162.0 grams of silicon is needed to produce 400.0 grams of chromium in the second reaction.

Key Concepts

Chemical Reactions

Chemical reactions are the processes where substances, known as reactants, are transformed into new substances called products. In our example, chromium is produced from two distinct reactions.
The first reaction uses chromium(III) oxide and aluminum as reactants, resulting in chromium metal and aluminum oxide as products. The second reaction replaces aluminum with silicon but still produces chromium metal along with silicon dioxide.
Key points about chemical reactions:

• They involve changes in chemical bonds.
• Reactants and products can differ in physical state.
• Energy can be absorbed or released.

Understanding these basics helps contextualize the stoichiometry of chromium production, which involves calculating the necessary amounts of reactants to yield a desired quantity of product.

Molar Mass

Molar mass is a core concept in chemistry that refers to the mass of one mole of a substance. Moles are basic counting units used in chemistry to express quantities of a substance, similar to how a dozen represents 12 items.
For instance, the molar mass of chromium (Cr) is 51.996 g/mol, indicating that one mole of chromium atoms weighs 51.996 grams. Aluminum (Al), another element in the reaction, has a molar mass of 26.982 g/mol.
Understanding molar masses is critical because they allow us to:

• Convert between mass and moles.
• Apply stoichiometric calculations to find amounts of reactants needed.

This concept becomes particularly useful in the calculation processes during chromium production.

Chromium Production

Chromium production via chemical reactions involves extracting chromium metal from chromium(III) oxide. In our exercises, two methods are covered:
1. **Using Aluminum**: Here, aluminum reduces chromium(III) oxide to produce chromium metal and aluminum oxide. 2. **Using Silicon**: Silicon also acts as a reducing agent, resulting in chromium metal and silicon dioxide.
The choice between aluminum and silicon as a reducing agent depends on various industrial conditions such as temperature, availability, and cost of materials. Each method involves precise stoichiometric calculations to determine the exact amount of reactant needed for producing a specific amount of chromium metal.

Chemical Equations

Chemical equations symbolize the processes of chemical reactions, illustrating reactants and products with stoichiometrically balanced formulas.
For the reactions involving chromium production:

• The first reaction equation is \(\mathrm{Cr}_{2} \mathrm{O}_{3} + 2 \mathrm{Al} \to 2 \mathrm{Cr} + \mathrm{Al}_{2} \mathrm{O}_{3} \).
• The second uses \(2 \mathrm{Cr}_{2} \mathrm{O}_{3} + 3 \mathrm{Si} \to 4 \mathrm{Cr} + 3 \mathrm{SiO}_{2} \).

These equations are critical because:

• They must be balanced to ensure the law of conservation of mass is respected.
• They highlight the stoichiometry, showcasing the molar relationships between reactants and products.

Beneath each formula lies a quantitative relationship that dictates how much of each reactant is needed to achieve a desired output of the product.