Problem 37 - Chemistry and Chemical Reactivity

Question

Aluminum dissolves readily in hot aqueous NaOH to give the aluminate ion, \(\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-},\) and \(\mathrm{H}_{2} .\) Write a balanced equation for this reaction. If you begin with \(13.2 \mathrm{g}\) of \(\mathrm{Al},\) what volume (in liters) of \(\mathrm{H}_{2}\) gas is produced when the gas is measured at \(22.5^{\circ} \mathrm{C}\) and a pressure of \(735 \mathrm{mm}\) Hg?

Step-by-Step Solution

Answer

The volume of hydrogen gas produced is approximately 18.9 liters.

Step 1: Write the Unbalanced Chemical Equation

First, determine the reactants and products. The chemical reaction involves aluminum (Al) reacting with sodium hydroxide (NaOH) to form the aluminate ion \( [\mathrm{Al}(\mathrm{OH})_4]^{-} \) and hydrogen gas \( \mathrm{H}_2 \). The unbalanced equation is: \[ \mathrm{Al} + \mathrm{NaOH} \rightarrow \mathrm{NaAl}(\mathrm{OH})_4 + \mathrm{H}_2 \]

Step 2: Balance the Chemical Equation

To balance the equation, we need to adjust the coefficients to have the same number of atoms of each element on both sides. The balanced equation is: \[ 2 \mathrm{Al} + 2 \mathrm{NaOH} + 6 \mathrm{H}_2\mathrm{O} \rightarrow 2 \mathrm{NaAl}(\mathrm{OH})_4 + 3 \mathrm{H}_2 \]

Step 3: Convert Grams of Aluminum to Moles

Calculate the number of moles of aluminum using its molar mass. The molar mass of \( \mathrm{Al} \) is approximately \( 26.98 \text{ g/mol} \). \[ \text{moles of Al} = \frac{13.2 \text{ g}}{26.98 \text{ g/mol}} \approx 0.489 \text{ mol} \]

Step 4: Use Stoichiometry to Find Moles of Hydrogen Gas

Use the balanced equation to find the moles of \( \mathrm{H}_2 \). According to the balanced equation, 2 moles of \( \mathrm{Al} \) produce 3 moles of \( \mathrm{H}_2 \). \[ \text{moles of } \mathrm{H}_2 = 0.489 \text{ mol Al} \times \frac{3 \text{ mol } \mathrm{H}_2}{2 \text{ mol Al}} = 0.734 \text{ mol } \mathrm{H}_2 \]

Step 5: Calculate Volume of Hydrogen Gas Produced

Apply the ideal gas law \( PV = nRT \) to find the volume of \( \mathrm{H}_2 \). Convert the pressure to atmospheres \( P = 735 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} \approx 0.967 \text{ atm} \), and the temperature to Kelvin \( T = 22.5 + 273.15 = 295.65 \text{ K} \). Use \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \).\[ V = \frac{nRT}{P} = \frac{0.734 \times 0.0821 \times 295.65}{0.967} \approx 18.9 \text{ L} \]

Key Concepts

Stoichiometry

Stoichiometry is an essential concept in chemistry that helps us predict the quantities of reactants and products involved in chemical reactions. Understanding stoichiometry involves using the balanced chemical equations to determine the proportions of substances that react or are produced.
In the case of aluminum reacting with sodium hydroxide, stoichiometry allows us to calculate how much hydrogen gas is produced from a given amount of aluminum. This process involves several key steps:

• Identifying the balanced equation, which shows the quantitative relationships between reactants and products.
• Using the coefficients in the equation to derive the mole ratios, thereby allowing conversion between moles of different substances.
• Applying these mole ratios to find out the amount of one substance when a specific amount of another is given.

For example, when 2 moles of aluminum react, 3 moles of hydrogen gas are produced. Thus, stoichiometry enables us to use known quantities to infer unknown quantities, ensuring that we can predict the outcomes of chemical reactions accurately.

Balanced Equations

Balanced chemical equations are like the backbone of stoichiometry. They provide the framework needed to calculate the relative amounts of reactants and products in a chemical reaction. To achieve a balanced equation, the same number and type of atoms must be present on both sides of the reaction arrow. This reflects the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.

When balancing equations, it is crucial to adjust coefficients — the numbers placed in front of the chemical formulas — rather than altering the chemical formulas themselves.

• Identify all the substances involved in the reaction.
• Write down their formulas and an initial unbalanced equation.
• Adjust the coefficients to ensure the same number of each type of atom appears on both sides.

In the example reaction, balancing the equation with coefficients ensures that the chemical reaction respects the conservation of atoms, resulting in an equal number of aluminum, sodium, oxygen, and hydrogen atoms on both sides.

Ideal Gas Law

The ideal gas law connects several state variables: pressure, volume, temperature, and the number of moles of a gas. It's an important tool in chemistry for calculating any one of these aspects if the others are known. The ideal gas law is expressed as \(PV = nRT\), where

• \(P\) is the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant (typically 0.0821 L atm K\(^{-1}\) mol\(^{-1}\)).
• \(T\) is the temperature in Kelvin.

To find the volume of hydrogen gas produced in this reaction, we rearrange the ideal gas law to solve for \(V\): \[ V = \frac{nRT}{P} \]
By converting all measurements to the appropriate units (pressure in atm, temperature in Kelvin), we ensure the accuracy of our calculations, thus obtaining the correct volume of the hydrogen gas based on the conditions provided.

Molar Mass

Molar mass is an essential concept that bridges the gap between the mass of a substance and the number of moles of that substance. Each element has a unique molar mass, expressed in grams per mole (g/mol), which can be found on the periodic table. This concept is crucial when converting between grams and moles, as it provides the conversion factor.
For example, in our exercise, we have aluminum with a molar mass of approximately 26.98 g/mol. If you start with 13.2 grams of aluminum, you can calculate the number of moles by dividing the mass by its molar mass:

• \(\text{moles of Al} = \frac{13.2 \text{ g}}{26.98 \text{ g/mol}} \approx 0.489 \text{ mol}\)

Understanding molar mass allows us to precisely determine how much of a substance we are working with, whether calculating amounts needed for reactions or amounts produced. This makes molar mass a fundamental building block in the study and application of chemistry.