Problem 39 - Chemistry The Molecular Science

Question

When a commercial drain cleaner containing sodium hydroxide and small pieces of aluminum is poured, along with water, into a clogged drain, this reaction occurs: $$ \begin{aligned} 2 \mathrm{Al}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \\ 2 \mathrm{NaAl}(\mathrm{OH})_{4}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g}) \end{aligned} $$ If \(6.5 \mathrm{~g} \mathrm{Al}\) and excess \(\mathrm{NaOH}\) are reacted, calculate the volume of \(\mathrm{H}_{2}\) gas produced at \(742 \mathrm{mmHg}\) and \(22.0^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The volume of \(\mathrm{H}_2\) gas produced is approximately 8.88 L.

Step 1: Convert Mass to Moles

First, we need to convert the mass of aluminum from grams to moles. The molar mass of aluminum (Al) is approximately 26.98 g/mol. Use the formula: \( \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \).Thus, \( \frac{6.5 \text{ g}}{26.98 \text{ g/mol}} \approx 0.241 \text{ moles of Al} \).

Step 2: Use Stoichiometry to Find Moles of \(\mathrm{H}_2\)

From the balanced chemical equation, we see that 2 moles of Al produce 3 moles of \(\mathrm{H}_2\). Therefore, we can set up a ratio: \( \frac{3 \text{ moles } \mathrm{H}_2}{2 \text{ moles } \mathrm{Al}} \).Now, calculate the moles of \(\mathrm{H}_2\) produced from the moles of Al: \( 0.241 \times \frac{3}{2} \approx 0.3615 \text{ moles of } \mathrm{H}_2 \).

Step 3: Use Ideal Gas Law to Find Volume

Use the ideal gas law equation \( PV = nRT \) to calculate the volume of \(\mathrm{H}_2\). First, convert the temperature to Kelvin: \( 22.0 + 273.15 = 295.15 \text{ K} \).Next, convert the pressure to atm: \( 742 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} \approx 0.9763 \text{ atm} \).Use these values in the ideal gas equation where \( R = 0.0821 \text{ L atm/mol K} \):\( V = \frac{nRT}{P} = \frac{0.3615 \times 0.0821 \times 295.15}{0.9763} \approx 8.88 \text{ L} \).

Key Concepts

Sodium Hydroxide Reaction

When sodium hydroxide (\( \text{NaOH} \)) is involved in chemical reactions, it often behaves as a strong base. In the given reaction, sodium hydroxide is used to dissolve aluminum (\( \text{Al} \)), which might seem unusual at first. The reaction:

• 2 \( \text{Al(s)} + 2 \text{NaOH(aq)} + 6 \text{H}_2\text{O(l)} \longrightarrow 2 \text{NaAl(OH)}_4\text{(aq)} + 3 \text{H}_2\text{(g)} \)

illustrates a redox process where aluminum acts as a reducing agent. In this reaction, aluminum is oxidized, and the water molecules provide the necessary oxygen. Sodium ions remain in solution, as spectator ions, not interfering with the core chemistry of the aluminum hydroxide complex formation. The product, hydrogen gas, is liberated as bubbles, visibly showing the reaction in action.
To sum up, knowing the role of water is critical, as it acts as both a solvent and a reactant, while sodium hydroxide facilitates the transition of aluminum into a soluble aluminate form.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It requires balanced chemical equations to understand how many moles of reactants are needed or produced.
In our reaction, we see that 2 moles of aluminum react with 2 moles of sodium hydroxide to produce 3 moles of hydrogen gas:

• 2 \( \text{Al} \) : 2 \( \text{NaOH} \) : 3 \( \text{H}_2 \)

This shows a mole ratio among the reactants and products. We used this ratio to calculate how much hydrogen gas is produced from a known amount of aluminum. By understanding stoichiometry, we could carry out the conversion detailed in step 2 of the solution, confirming that 0.241 moles of aluminum produce approximately 0.3615 moles of hydrogen gas.
Stoichiometry is essential because it helps chemists predict the amounts of substances consumed and formed in a reaction, ensuring none of the reactants go to waste.

Moles to Volume Conversion

Converting moles to volume is a common task in chemistry, especially when dealing with gases. The Ideal Gas Law is the go-to formula for this conversion, defined as:\[ PV = nRT \]where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the universal gas constant
• \( T \) is the temperature in Kelvin

In our problem, we were tasked with finding the volume of the hydrogen gas at a given pressure and temperature. By converting temperature to Kelvin and pressure to atmospheres, we plugged these values into the Ideal Gas Law to determine the volume. Using the calculated moles of hydrogen (\( n = 0.3615 \) moles), the calculated volume came to about 8.88 liters. This process shows how understanding concepts like the gas constant and converting units can help predict real-world data from abstract equations, which is the heart of using the Ideal Gas Law effectively.