Problem 107 - Chemistry

Question

Some commercial drain cleaners contain a mixture of sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs: $$ \begin{array}{r} 2 \mathrm{NaOH}(a q)+2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \\ 2 \mathrm{NaAl}(\mathrm{OH})_{4}(a q)+3 \mathrm{H}_{2}(g) \end{array} $$ The heat generated in this reaction helps melt away obstructions such as grease, and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of \(\mathrm{H}_{2}\) formed at \(23^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) if \(3.12 \mathrm{~g}\) of \(\mathrm{Al}\) are treated with an excess of \(\mathrm{NaOH}\).

Step-by-Step Solution

Answer

The volume of \(H_{2}\) formed is 4.2 liters.

Step 1: Determine Moles of Al

First, the number of moles of aluminum (\(Al\)) treated should be determined. This can be done by dividing the mass of the aluminum by its molar mass. The atomic mass of aluminum is approximately \(26.98 \, g/mol\), so \(3.12 \, g\) of \(Al\) corresponds to \(\frac{3.12 \, g}{26.98 \,g/mol}\) which equals to \(0.116 \, mol\).

Step 2: Stoichiometric Calculation

The stoichiometric relationship between \(Al\) and \(H_2\) is 2:3 in the given equation. So for every 2 mol of \(Al\), 3 mol of \(H_2\) are formed. Thus, the amount of \(H_2\) gas formed is \((0.116 \, mol) \times \frac{3}{2} = 0.174 \, mol\).

Step 3: Calculate Volume using Ideal Gas Law

Next, to find the volume of \(H_2\) gas formed, the ideal gas law can be used. The ideal gas law states that \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. Given \(P = 1.00 \, atm\), \(n = 0.174 \, mol\), \(R = 0.0821 \, L.atm/K.mol\), and \(T = 23^{\circ}C = 296 \, K\). Solving the equation for \(V\) yields \(V = \frac{nRT}{P} = \frac{(0.174 \, mol)(0.0821 \, L.atm/K.mol)(296 \, K)}{1.00 \, atm} = 4.2 \, L\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that connects the pressure, volume, temperature, and amount (in moles) of a gas. This relationship is shown in the equation \(PV = nRT\), where:

• \(P\) stands for pressure, measured in atmospheres (atm).
• \(V\) represents volume, typically in liters (L).
• \(n\) is the number of moles of gas, an important measure of quantity.
• \(R\) is the ideal gas constant, \(0.0821 \, L.atm/K.mol\), providing proportionality across the equation.
• \(T\) signifies temperature in Kelvin, important for maintaining the equation's accuracy.

To use the Ideal Gas Law, convert the temperature in Celsius to Kelvin by adding 273.15. For example, \(23^{\circ}C = 296\, K\). With everything in proper units, the law helps determine any one of these variables if the others are known.
It's especially useful in problems involving reactions that release or absorb gases, helping predict gas volumes.

Moles of Aluminum

In chemistry, moles are the standard unit for measuring the amount of a substance. To figure out how much aluminum is involved in a reaction, we need to calculate its moles. This process uses the formula: \[moles = \frac{\text{mass of substance}}{\text{molar mass of the substance}}\]For aluminum, with a molar mass of approximately \(26.98 \, g/mol\), we can determine the moles from a given mass. In our example, \(3.12 \, g\) of aluminum converts to moles by the equation: \[\frac{3.12 \, g}{26.98 \, g/mol} \approx 0.116 \, mol\]
Understanding this conversion from grams to moles is a pivotal step in stoichiometry, enabling us to participate in the reaction's calculations accurately.

Reaction Equation

Chemical reactions are depicted using balanced chemical equations, showcasing the reactants and products involved. In this context, our equation is:\[2 \mathrm{NaOH}(aq) + 2 \mathrm{Al}(s) + 6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{NaAl(OH)}_4(aq) + 3 \mathrm{H}_2(g)\] This representation provides vital information about stoichiometry, the quantitative relationship of reactants and products. In this case, the reaction shows that:

• 2 moles of aluminum (\(Al\)) react with sodium hydroxide (\(NaOH\)) and water to produce 3 moles of hydrogen gas (\(H_2\)).
• By interpreting the balanced equation, you can determine how moles of one substance relate to moles of another.

This stoichiometric consistency allows calculation of gas production based on the moles of reactants initially present.

Volume of Gas

Determining gas volume is crucial in understanding its practical application, such as in our reaction where hydrogen gas is released. By using the ideal gas law, we can uncover this volume. For instance, knowing:

• \(n = 0.174 \, mol\) of \(H_2\)
• \(P = 1.00 \, atm\)
• \(R = 0.0821 \, L.atm/K.mol\)
• \(T = 296 \, K\)

we substitute these into \(PV = nRT\), solving for \(V\): \[V = \frac{(0.174 \, mol)(0.0821 \, L.atm/K.mol)(296 \, K)}{1.00 \, atm} = 4.2 \, L\]Thus, the volume of hydrogen gas produced is \(4.2 \, liters\). This calculation aids in anticipating how much space the gas will occupy post-reaction, highlighting its significant role in chemistry and real-world applications.