Problem 40 - Chemistry and Chemical Reactivity

Question

"Aerated" concrete bricks are widely used building materials. They are obtained by mixing gas-forming additives with a moist mixture of lime, cement, and possibly sand. Industrially, the following reaction is important: \(-2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$3 \mathrm{CaO} \cdot \mathrm{Al}_{2} \mathrm{O}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g})$$ Assume the mixture of reactants contains 0.56 g of \(\mathrm{Al}\) (as well as excess calcium hydroxide and water) for each brick. What volume of hydrogen gas do you expect at \(26^{\circ} \mathrm{C}\) and a pressure of \(745 \mathrm{mm}\) Hg?

Step-by-Step Solution

Answer

0.780 L of hydrogen gas is produced.

Step 1: Molar Mass of Aluminum

First, find the molar mass of aluminum (Al). Aluminum has an atomic mass of \(27\ \text{g/mol}\).

Step 2: Moles of Aluminum

Calculate the moles of aluminum using its mass and molar mass. \[\text{Moles of } \mathrm{Al} = \frac{0.56\ \text{g}}{27\ \text{g/mol}} \approx 0.02074\ \text{mol}\]

Step 3: Reaction Stoichiometry

From the balanced chemical equation, \(-2 \mathrm{Al} + 3 \mathrm{Ca(OH)}_2 + 6 \mathrm{H}_2\mathrm{O} \rightarrow 3 \mathrm{CaO} \cdot \mathrm{Al}_2\mathrm{O}_3 \cdot 6 \mathrm{H}_2\mathrm{O} + 3 \mathrm{H}_2\),\ we see each 2 moles of Al produce 3 moles of \(\mathrm{H}_2\).

Step 4: Moles of Hydrogen Gas

Use the stoichiometry of the reaction to find moles of hydrogen gas. For every 2 moles of aluminum, 3 moles of hydrogen gas (\(\mathrm{H}_2\)) are produced.\[\text{Moles of } \mathrm{H}_2 = 0.02074\ \text{mol Al} \times \frac{3\ \text{mol } \mathrm{H}_2}{2\ \text{mol Al}} = 0.03111\ \text{mol } \mathrm{H}_2\]

Step 5: Ideal Gas Law

Convert the given temperature to Kelvin by adding 273 to the Celsius temperature:\[T(K) = 26 + 273 = 299\ \text{K}\]Convert pressure from mm Hg to atm:\[P = \frac{745\ \text{mm Hg}}{760\ \text{mm Hg/atm}} \approx 0.98026\ \text{atm}\]Now use the ideal gas law \(PV = nRT\) to calculate the volume of hydrogen gas. Use \(R = 0.0821\ \text{L atm/mol K}\):\[V = \frac{nRT}{P} = \frac{0.03111\ \text{mol} \times 0.0821\ \text{L atm/mol K} \times 299\ \text{K}}{0.98026\ \text{atm}} \approx 0.780\ \text{L}\]

Step 6: Volume of Hydrogen Gas

Calculate the final volume of hydrogen gas using the information from the above steps. This calculation places \(V = 0.780\ \text{L}\) of hydrogen gas produced at \(26^\circ \text{C}\) and \(745\ \text{mm Hg}\).

Key Concepts

Understanding Stoichiometry

Stoichiometry is a concept in chemistry that involves calculating the relative quantities of reactants and products involved in a chemical reaction. It is crucial to understand this concept to solve problems like the one about the aerated concrete bricks. Stoichiometry uses the balanced chemical equation to determine the ratios of moles of reactants and products. For instance, in the reaction between aluminum and calcium hydroxide forming hydrogen gas:

• The balanced equation is: \[-2 \mathrm{Al} + 3 \mathrm{Ca(OH)}_2 + 6 \mathrm{H}_{2}\mathrm{O} \rightarrow 3 \mathrm{CaO} \cdot \mathrm{Al}_2\mathrm{O}_3 \cdot 6 \mathrm{H}_2\mathrm{O} + 3 \mathrm{H}_2\]
• This tells us that 2 moles of Al produce 3 moles of \(\mathrm{H}_2\).
• The stoichiometric ratio is used to calculate the amount of products formed given the amount of reactants available.

These calculations help us predict how much of a product will be formed in a reaction, just like predicting the volume of hydrogen gas in the given problem.

Delving into Chemical Reactions

Chemical reactions involve the transformation of substances into new products through breaking and forming chemical bonds. In the given exercise, the reaction transforms aluminum and other compounds into a new product with the release of hydrogen gas. There are several key aspects to understand in chemical reactions:

• Reactants and Products: Reactants are substances that start a chemical reaction. In this problem, aluminum, calcium hydroxide, and water are the reactants.
• Balancing Reactions: A balanced chemical equation ensures that the number of atoms of each element is the same on both sides. This balancing is vital for stoichiometry calculations.
• Types of Reactions: Such reactions often involve synthesis and decomposition. Here, the synthesis reaction forms a compound while releasing gas.

Understanding the type and nature of a chemical reaction allows chemists to predict and control the outcomes effectively, like in the industrial use of aerated concrete bricks.

The Importance of Molar Mass

The molar mass is a fundamental concept that relates the mass of a substance to the amount in moles. It is essential for converting grams of a material to moles, which is necessary for stoichiometric calculations. Molar mass is expressed in grams per mole (g/mol). Here's how it applies to aluminum in our scenario:

• Aluminum has a molar mass of 27 g/mol.
• To find the moles of aluminum, divide its mass by its molar mass: \[\text{Moles of } \mathrm{Al} = \frac{0.56 \text{ g}}{27 \text{ g/mol}} \approx 0.02074 \text{ mol}\]
• This calculation allows us to determine how many moles of reactant are available to produce a certain amount of product.

Using molar mass, you can bridge the gap between the mass of a substance and the stoichiometric calculations required to predict the outcome of a chemical reaction.

Exploring Gas Laws

Gas laws are essential for predicting the behavior of gases under different conditions of temperature, pressure, and volume. In the exercise, we use the Ideal Gas Law to find the volume of hydrogen gas produced.Here are the key elements of gas laws used in the problem:

• Ideal Gas Law: This law is represented as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.
• Calculating Volume: Once the moles of \(\mathrm{H}_2\) are known, convert temperature to Kelvin and pressure to atmospheres, then rearrange the Ideal Gas Law to solve for volume: \[V = \frac{nRT}{P}\]
• The calculation involves using the known moles of hydrogen, the gas constant \(R = 0.0821\) L atm/mol K, the converted temperature, and pressure to find the expected volume of gas produced.

Understanding gas laws is crucial as it allows the prediction of how changes in conditions affect gaseous substances, like the volume of hydrogen gas in this industrial application.