Problem 66 - Chemistry and Chemical Reactivity

Question

You can dissolve an aluminum soft drink can in an aqueous base such as potassium hydroxide. $$2 \mathrm{Al}(\mathrm{s})+2 \mathrm{KOH}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{KAl}(\mathrm{OH})_{4}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})$$ 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

No aluminum will remain; 11.87 g of \(\text{KAl(OH)}_4\) is produced.

Step 1: Identify the Limiting Reactant

To find the limiting reactant, we need to determine how many moles of each reactant are present. The molar mass of aluminum Al is approximately 26.98 g/mol. Calculate the moles of aluminum: \(\frac{2.05\, \text{g Al}}{26.98\, \text{g/mol}} = 0.076\, \text{mol Al}\). The potassium hydroxide solution has a molarity of 1.35 M and a volume of 185 mL. Convert the volume to liters: 185 mL = 0.185 L. Calculate the moles of KOH: \(1.35\, \text{mol/L} \times 0.185\, \text{L} = 0.25\, \text{mol KOH}\). According to the balanced equation, 2 moles of Al react with 2 moles of KOH. Compare the moles available to find the limiting reactant.

Step 2: Determine the Limiting Reactant

From Step 1, we have 0.076 mol of Al and 0.25 mol of KOH. According to the stoichiometry of the reaction, 2 mol of Al reacts with 2 mol of KOH. Since \(0.076\, \text{mol Al} < 0.25\, \text{mol KOH}\), Al is the limiting reactant.

Step 3: Calculate the Mass of Product (Potassium Aluminum Tetrahydroxide)

We see from the balanced equation that 2 moles of Al produce 2 moles of \(\text{KAl(OH)}_4\). Therefore, 0.076 moles of Al will produce 0.076 moles of \(\text{KAl(OH)}_4\). The molar mass of \(\text{KAl(OH)}_4\) is approximately \(M(\text{K}) = 39.10\, \text{g/mol}, M(\text{Al}) = 26.98\, \text{g/mol}, M(\text{O}) = 16.00\, \text{g/mol}, M(\text{H}) = 1.01\, \text{g/mol}\). The total molar mass is: \(39.10 + 26.98 + 4(16.00 + 1.01) = 156.12\, \text{g/mol}\). Calculate the mass of \(\text{KAl(OH)}_4\) formed: \(0.076\, \text{mol} \times 156.12\, \text{g/mol} = 11.87\, \text{g}\).

Step 4: Check for Remaining Reactants

We have established that Al is the limiting reactant. Once the reaction has gone to completion, no Al will remain because it has all reacted. Therefore, only KOH will be in excess.

Key Concepts

Limiting Reactant

When it comes to chemical reactions, identifying the limiting reactant is crucial. This is because the limiting reactant is the substance that will be consumed first, stopping the reaction from continuing further. Think of it like baking cookies: if you run out of sugar first, you can't make more cookies, even if you have plenty of flour and butter left.
To find the limiting reactant, calculate the moles of each reactant present. Use the balanced chemical equation to compare the mole ratios. The reactant that is all used up first is your limiting reactant. In the given problem, aluminum (Al) ends up being the limiting reactant because when compared to potassium hydroxide (KOH), it is available in fewer moles.

Molar Mass Calculation

Calculating molar mass is essential for understanding how much of a product you will get from a chemical reaction. Molar mass is the mass of one mole of a substance and is typically expressed in grams per mole (g/mol).

Here's how to calculate the molar mass for complicated molecules like potassium aluminum tetrahydroxide (KAl(OH) _4 ). First, gather the molar masses of each element involved:

• Potassium (K): 39.10 g/mol
• Aluminum (Al): 26.98 g/mol
• Oxygen (O): 16.00 g/mol
• Hydrogen (H): 1.01 g/mol

Using these values, add them according to the molecular formula: K + Al + 4(O + H). This provides a comprehensive approach to determining how much of each product can be produced and is crucial for the stoichiometry calculations.

Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. It shows the starting substances (reactants) and the resulting substances (products), along with their respective amounts. Chemical equations are essential for visualizing what happens during a reaction.
The balanced equation for our problem is:\[2 \text{Al} + 2 \text{KOH} + 6 \text{H}_2\text{O} \rightarrow 2 \text{KAl(OH)}_4 + 3 \text{H}_2\]

Balancing a chemical equation is vital because it ensures the law of conservation of mass is observed. This means there are the same number of each type of atom on both sides of the reaction. Balancing provides the correct stoichiometry, which we use for calculating the limiting reactant and product formation.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative study of reactants and products in a chemical reaction. It allows us to predict how much product will form from given quantities of reactants or how much of a reactant is required to make a certain amount of product.

Start with the balanced chemical equation, which indicates the ratio in which reactants combine and products form. Using the equation provided, each mole of aluminum reacts with one mole of KOH to produce one mole of potassium aluminum tetrahydroxide. From our scenario, 0.076 moles of Al will yield 0.076 moles of product.

• The balanced equation provides a roadmap.
• Using molar ratios helps convert moles of one substance into moles of another.
• This process highlights how balanced equations serve as the foundation for predicting reactant consumption and product formation.

With this knowledge, stoichiometry serves as a vital tool for any chemical calculations.