Problem 66 - Chemistry and Chemical Reactivity

Question

You can dissolve an aluminum soft-drink can in an aqueous base such as potassium hydroxide. \(2 \mathrm{Al}(\mathrm{s})+2 \mathrm{KOH}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow\) $$2 \mathrm{KAl}(\mathrm{OH})_{4}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})$$ If you place \(2.05 \mathrm{g}\) of aluminum in a beaker with \(185 \mathrm{mL}\) of \(1.35 \mathrm{M} \mathrm{KOH},\) will any aluminum remain? What mass of \(\mathrm{KAl}(\mathrm{OH})_{4}\) is produced?

Step-by-Step Solution

Answer

Aluminum is the limiting reactant, producing 11.93 g of KAl(OH)_4, with no aluminum remaining.

Step 1: Determine Moles of Reactants

First, calculate the moles of aluminum and potassium hydroxide. For aluminum, use its molar mass of about 26.98 g/mol:\[ \text{Moles of } \mathrm{Al} = \frac{2.05 \text{ g}}{26.98 \text{ g/mol}} = 0.076 \text{ moles} \]For potassium hydroxide (KOH), use the given molarity and volume to calculate moles:\[ \text{Moles of KOH} = 1.35 \text{ M} \times 0.185 \text{ L} = 0.24975 \text{ moles} \]

Step 2: Identify the Limiting Reactant

Using the reaction stoichiometry (2 mol Al : 2 mol KOH), you need 1 mole of aluminum for 1 mole of KOH. Therefore: - Moles of KOH needed for 0.076 moles of aluminum = 0.076 moles. With 0.24975 moles of KOH available, aluminum is the limiting reactant because you have more than enough KOH to react with the available aluminum.

Step 3: Calculate Product Formation

The reaction shows that 2 moles of aluminum produce 2 moles of KAl(OH)_4_. Thus, moles of KAl(OH)_4_ formed are equal to the moles of aluminum used.\[ \text{Moles of } \mathrm{KAl(OH)}_4 = 0.076 \text{ moles} \]Now, calculate the mass using its molar mass (157.1 g/mol):\[ \text{Mass of } \mathrm{KAl(OH)}_4 = 0.076 \text{ moles} \times 157.1 \text{ g/mol} = 11.934 \text{ g} \]

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is entirely consumed first, dictating the amount of product formed. To identify the limiting reactant, you compare the initial amounts of reactants to the stoichiometric ratios in the balanced chemical equation.
For instance, in the reaction dissolving aluminum in potassium hydroxide, the stoichiometry is 2 moles of aluminum to 2 moles of potassium hydroxide. After determining the moles of each reactant, we found 0.076 moles of aluminum and 0.24975 moles of KOH. We only need 0.076 moles of KOH to react with our aluminum, leaving aluminum as the limiting reactant since it's used up first.

• Key point: The limiting reactant determines the maximum product yield.
• In our reaction, aluminum limits the product formation.

Recognizing the limiting reactant ensures precise predictions of how much product will form and allows for efficient resource usage in real-world applications.

Stoichiometry

Stoichiometry is the chemistry concept that helps us understand the quantitative relationships between reactants and products in a chemical reaction. It allows us to use balanced chemical equations to predict how much of each substance is involved.
When given the balanced equation \[2 \mathrm{Al}(\mathrm{s})+2 \mathrm{KOH}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{KAl}(\mathrm{OH})_{4}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})\] we see a 2:2 mole ratio between aluminum and potassium hydroxide.

• This tells us, for every 2 moles of aluminum, 2 moles of KOH are needed for the reaction.
• This ratio is crucial for identifying which reactant runs out first and dictates the reaction's extent.

Understanding stoichiometry is essential for solving real-world problems in chemistry by ensuring reactions are both balanced and efficient.

Molar Mass

Molar mass is a fundamental concept in understanding the mass-to-mole conversion for substances in stoichiometry. It is defined as the mass of one mole of a given molecule or atom, typically expressed in grams per mole (g/mol).
In our scenario with aluminum and KOH, we calculated the moles using their respective molar masses:

• Aluminum: about 26.98 g/mol, crucial for converting the mass of aluminum given in grams to moles, allowing us to then use stoichiometry effectively.
• KOH: The molar mass comes into play when calculating moles from a solution's molarity. This enables us to strategize the moles of reactants needed for the reaction.

The concept of molar mass makes abstract quantities like atoms and molecules practically measurable, bridging the gap between the microscopic and macroscopic worlds in chemistry.

Product Formation

Once you've identified the limiting reactant and calculated the necessary stoichiometric ratios, you can determine the amount of product formed in a reaction. This is known as product formation.
In the reaction between aluminum and potassium hydroxide, once the stoichiometry is determined, we see that each mole of aluminum results in one mole of potassium aluminum hydroxide (KAl(OH) \(_4\)), making it simple to predict product formation. The final step involves finding the mass of the product formed. We use the molar mass of the product (KAl(OH) \(_4\), 157.1 g/mol) to calculate:

• 0.076 moles of aluminum produces 0.076 moles of KAl(OH) \(_4\).
• Multiplying by the molar mass gives approximately 11.934 grams of KAl(OH) \(_4\) produced.

This step not only allows understanding of theoretical yield but also helps compare with actual experimental results.