Problem 59 - Chemistry An Atoms First Approach

Question

What mass of solid aluminum hydroxide can be produced when 50.0 mL of \(0.200 M\) Al(NO_j) is added to 200.0 mL of \(0.100 M\) KOH?

Step-by-Step Solution

Answer

Approximately 0.520 grams of solid aluminum hydroxide can be produced in this reaction.

Step 1: Determine the moles of each reactant

To determine the moles of each reactant, the volume (V) of the solutions and their molar concentration (c) should be used. The formula is: moles = V (in L) × c (in mol/L) For aluminum nitrate: moles of Al(NO₃)₃ = (50.0 mL × 0.200 mol/L) / 1000 moles of Al(NO₃)₃ = \( 0.0100 \) mol For potassium hydroxide: moles of KOH = (200.0 mL × 0.100 mol/L) / 1000 moles of KOH = \( 0.0200 \) mol

Step 2: Identify the limiting reactant

In order to identify the limiting reactant, we can compare the mole ratio from the balanced chemical equation and the mole ratio of the reactants we are given. From the chemical equation, we know that: 1 mol Al(NO₃)₃ : 3 mol KOH Now, divide the moles of each reactant by their respective coefficients in the balanced chemical equation: For Al(NO₃)₃: \( \frac{0.0100 \text{ mol}}{1} = 0.0100 \) For KOH: \( \frac{0.0200 \text{ mol}}{3} = 0.00667 \) Since 0.00667 < 0.0100, KOH is the limiting reactant.

Step 3: Calculate the moles of Al(OH)₃ produced

Now that we know KOH is the limiting reactant, we can calculate the moles of aluminum hydroxide formed. From the balanced chemical equation, the mole ratio is 1 mol Al(NO₃)₃ : 1 mol Al(OH)₃. moles of Al(OH)₃ = 0.00667 (from KOH) × 1 (ratio of Al(OH)₃ to KOH) moles of Al(OH)₃ = \( 0.00667 \) mol

Step 4: Convert moles of Al(OH)₃ to mass

Finally, we will find the mass of aluminum hydroxide by converting the moles of Al(OH)₃ to grams using its molar mass. The molar mass of Al(OH)₃ is: 27.00 g/mol (Al) + 3 × (16.00 g/mol + 1.00 g/mol (OH component)) = 78.00 g/mol Now, we can convert the moles of Al(OH)₃ to grams: mass of Al(OH)₃ = \( 0.00667 \) mol × 78.00 g/mol mass of Al(OH)₃ ≈ 0.520 g So, approximately 0.520 grams of solid aluminum hydroxide can be produced in this reaction.

Key Concepts

Limiting Reactant

In any chemical reaction, not all reactants are consumed equally. The limiting reactant is the one that is used up first, thus determining how much product can be formed.
It's like baking cookies where you might run out of sugar while still having flour left. To find out which reactant limits the product formation, we need to look at the balanced chemical equation and compare the mole ratios.In this practice problem, we have aluminum nitrate (\( \text{Al(NO}_3)_3 \)) and potassium hydroxide (\( \text{KOH} \)). According to the balanced equation:

• 1 mole of \( \text{Al(NO}_3)_3 \)
• 3 moles of \( \text{KOH} \)

The mole ratio lets us know where we start. By comparing the initial moles given in the problem with the balanced equation's ratios, we figure out that \( \text{KOH} \) is the limiting reactant because it is in shorter supply relative to \( \text{Al(NO}_3)_3 \).
With limiting reactants, knowing which one is consumed first helps us calculate the maximum amount of product that can be formed.

Mole Ratio

A balanced chemical equation gives us the mole ratio of the reactants and products, key to determining which reactant is limiting. The mole ratio tells us how many moles of each reactant are needed and how many moles of each product are formed.
It is the bridge that connects reactants to products in stoichiometry.For the reaction involving \( \text{Al(NO}_3)_3 \) and \( \text{KOH} \), the balanced equation shows:

• 1 mole of \( \text{Al(NO}_3)_3 \) reacts with 3 moles of \( \text{KOH} \)
• Producing 1 mole of aluminum hydroxide (\( \text{Al(OH)}_3 \))

The mole ratio is 1:3:1, meaning for every mole of \( \text{Al(NO}_3)_3 \) used, three moles of \( \text{KOH} \) are required for complete reaction to produce one mole of \( \text{Al(OH)}_3 \).
This proportionality is vital to understanding how much of each substance is needed or will be produced in a reaction, making it an important concept in stoichiometry.

Molar Mass Calculation

After identifying the number of moles of substances involved, the next step is to convert moles into grams because mass is a more tangible measure.
Enter the concept of molar mass, which is the mass of one mole of a substance.For aluminum hydroxide (\( \text{Al(OH)}_3 \)), the molar mass is calculated as:

• Aluminum (\( \text{Al} \)): 27.00 g/mol
• Three Hydroxide ions (\( \text{OH}^- \)): 3 × (16.00 g/mol + 1.00 g/mol) for oxygen and hydrogen.
• Total: 78.00 g/mol

After determining the moles of aluminum hydroxide produced, this molar mass allows us to calculate its actual mass in grams.
Simply multiply the number of moles by the molar mass: 0.00667 moles \( \times \) 78.00 g/mol = approx. 0.520 grams of aluminum hydroxide.
This conversion is crucial in translating chemical reactions into meaningful, real-world quantities.

Chemical Reaction

A chemical reaction involves rearranging the atoms in reactants to form new products. In our example, aluminum nitrate reacts with potassium hydroxide to form aluminum hydroxide and potassium nitrate.
This process can be described by a balanced chemical equation, which ensures that the same number and type of atoms are present both before and after the reaction. In simpler terms, think of a chemical reaction as a chemical recipe. Each reactant acts as an ingredient, combining in specific amounts (mole ratios) established by the balanced equation to create the product.
A balanced reaction is critical because it necessitates that all atoms are conserved, aligning with the Law of Conservation of Mass which states that matter is neither created nor destroyed.
Understanding the reactants and products of a chemical equation helps us determine what substances interact and what amounts are needed or produced, giving insight into the stoichiometric relationships involved.