Problem 105 - General Chemistry: Principles and Modern Applications

Question

A drop \((0.05 \mathrm{mL})\) of \(12.0 \mathrm{M} \mathrm{HCl}\) is spread over a sheet of thin aluminum foil. Assume that all the acid reacts with, and thus dissolves through, the foil. What will be the area, in \(\mathrm{cm}^{2}\), of the cylindrical hole produced? (Density of \(\mathrm{Al}=2.70 \mathrm{g} / \mathrm{cm}^{3} ;\) foil thickness \(=0.10 \mathrm{mm} .)\) \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})\)

Step-by-Step Solution

Answer

The area of the cylindrical hole produced in the aluminum foil is 0.2 cm².

Step 1: Calculate the number of moles of HCl

Molarity (M) is given by the number of moles of solute per litre of solution. Given that, 12.0 M HCl means there are 12.0 moles of HCl in 1 liter or 1000 mL of solution.So, the number of moles of HCl in 0.05 mL can be calculated using the equation: \[\text{{number of moles of HCl}} = \text{{Molarity}} \times \text{{Volume in litres}}\]Or,\[\text{{number of moles of HCl}} = 12.0M \times 0.05 mL \times \frac{1L}{1000 mL}= 0.0006 \text{{moles}}\]

Step 2: Use stoichiometry to determine the volume of Al

Looking at the balanced chemical equation: \[2Al(s) + 6HCl(aq) \rightarrow 2AlCl3(aq) + 3H2(g)\]we can see that 6 moles of HCl react with 2 moles of Al. Hence 0.0006 moles of HCl will react with:\[\text{{moles of Al}} = \frac{2}{6} \times 0.0006 \text{{ moles of HCl}} = 0.0002 \text{{ moles of Al}}\]Now, using density formula, \(density = \frac{mass}{volume}\), the volume of Al can be calculated as:\[\text{{volume of Al}} = \frac{\text{{moles of Al}} \times \text{{molar mass of Al}}}{\text{{density of Al}}} \]Substituting the values, \[\text{{volume of Al}} = \frac{0.0002 \text{{moles of Al}} \times 26.98 \text{{g/mol}}}{2.70 \text{{g/cm}}^{3}} = 0.002 \text{{cm}}^{3}\]

Step 3: Calculate the area of the hole

The aluminum foil is shown as a cylinder so the area of the hole (base of the cylinder) can be found using volume formula \(V = πr^{2}h\), where r is the radius and h is the height (thickness of the foil). Here, volume \(V = 0.002 \text{{cm}}^{3}\) and thickness \(h = 0.10 \text{{mm}} = 0.01 \text{{cm}}\).So, the area of the hole can be calculated as: \[Area = \frac{V}{h} = \frac{0.002 \text{{cm}}^{3}}{0.01 \text{{cm}}} = 0.2 \text{{cm}}^{2}\]

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In this process, bonds are broken, rearranged, and new bonds are formed to create different substances. The reaction between aluminum (\( \text{Al} \)) and hydrochloric acid (\( \text{HCl} \)) is a perfect example to illustrate stoichiometry. The given balanced chemical equation is \( 2 \text{Al(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{AlCl}_3\text{(aq)} + 3 \text{H}_2\text{(g)} \), showing that it takes 6 moles of hydrochloric acid to react with 2 moles of aluminum to produce 2 moles of aluminum chloride and 3 moles of hydrogen gas.

This exercise requires using stoichiometry to relate the amount of hydrochloric acid reacting to the quantity of aluminum involved. Stoichiometry is key for predicting the outcome of reactions, ensuring reactions are balanced, and understanding the proportions used in chemical reactions.

Molarity

Molarity is the concentration of a solution expressed as the number of moles of solute per liter of solution. It is a crucial concept for understanding solutions in chemistry. In the exercise, the molarity of hydrochloric acid is given as 12.0 M, meaning there are 12 moles of \( \text{HCl} \) per liter of solution.

To find how many moles of \( \text{HCl} \) are contained in a small volume, we use the formula:

• \( \text{Moles of HCl} = \text{Molarity} \times \text{Volume in liters} \)

The solution for a drop of 0.05 mL of \( \text{HCl} \) indicates that the number of moles is 0.0006, an essential step for further stoichiometric calculations needed in this solution.

Density

Density is defined as mass per unit volume. It is an important property in calculating the volume occupied by a given mass of a substance. For aluminum with a density of 2.70 g/cm³, we can use the density formula:

• \( \text{Density} = \frac{\text{mass}}{\text{volume}} \)

This formula helps us in calculating the volume of aluminum required for the reaction. For example, if we know the number of moles and the molar mass, we can determine the mass of aluminum needed. The volume is then derived using the density, thereby providing insights into how much space the aluminum occupies after reacting with \( \text{HCl} \). This is fundamental in the volume and area calculation of the aluminum foil hole.

Volume Calculation

Volume calculations in this context are essential for determining the size of the hole created in the aluminum foil. By using the formula for the volume of a cylinder \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height (or thickness of the foil), we find the base area of the cylindrical hole.

The given volume \( V = 0.002 \text{ cm}^3 \) and thickness \( h = 0.01 \text{ cm} \) are used in the step:

• \( \text{Area} = \frac{\text{Volume}}{\text{Height}} \)

which results in an area of 0.2 cm² for the hole. Such calculations are direct applications of physical properties in chemical reactions, facilitating understanding of how substances interact and measure geometrically after a reaction. Understanding the relationship between volume, thickness, and area is crucial for comprehensively addressing reaction impacts on materials.