Question

An experiment calls for a \(0.4500 \mathrm{M}\) solution of aluminum carbonate. The storeroom has only \(32.00 \mathrm{~g}\) of the solid. (a) What is the maximum volume of solution that can be prepared? (b) The reaction in the experiment requires \(0.1450 \mathrm{~mol}\) of carbonate ion. How many milliliters of this solution are required to provide 0.1450 mol?

Short answer

(b) What is the volume of the solution required to provide 0.1450 mol of carbonate ion? Answer: (a) The maximum volume of the 0.4500 M solution of aluminum carbonate that can be prepared is ___ L or ___ mL. (b) The volume of the solution required to provide 0.1450 mol of carbonate ion is ___ L or ___ mL.

Step-by-step solution

(Step 1: Calculate the molar mass of aluminum carbonate)

To calculate the moles of aluminum carbonate, we first need to determine its molar mass. Aluminum carbonate (Al2(CO3)3) contains 2 aluminum atoms, 3 carbonate groups (CO3), and hence 3 carbon atoms and 9 oxygen atoms. Using the atomic masses of Al, C, and O, we can find the molar mass of aluminum carbonate: Molar mass = 2(Al) + 3(C) + 9(O) Where Al = 26.98 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol.

(Step 2: Calculate the moles of aluminum carbonate from its mass)

Given the mass of aluminum carbonate (32.00 g), we can now calculate the moles using the molar mass determined in step 1: Moles of aluminum carbonate = (mass of aluminum carbonate)/(molar mass of aluminum carbonate)

(Step 3: Calculate the maximum volume of the 0.4500 M solution)

To find the maximum volume of the 0.4500 M solution, we will use the calculated moles of aluminum carbonate and the given molarity: Molarity = (moles of solute)/(volume of solution in liters) Rearrange this formula to get the volume of solution: Volume of solution in liters = (moles of solute)/(molarity)

(Step 4: Calculate the volume of solution required to provide 0.1450 mol of carbonate ion)

For this step, we will need to find the relationship between aluminum carbonate and carbonate ion. Since 2 moles of aluminum carbonate provides 3 moles of carbonate ion, we can use this ratio to calculate moles of aluminum carbonate needed: moles of aluminum carbonate = (0.1450 mol carbonate ion) x (2 moles of aluminum carbonate/3 moles of carbonate ion) Now that we have the moles of aluminum carbonate needed, we can calculate the volume of the solution by using the given molarity as in step 3: Volume of solution in liters = (moles of aluminum carbonate)/(molarity) Finally, convert the volume from liters to milliliters (1 L = 1000 mL) to obtain the required volume in milliliters.

Key concepts

Molar Mass Determination

The molar mass of a substance represents the mass of one mole of its particles, be it atoms, molecules, or formula units. It's a critical factor in many chemistry calculations, especially in solution preparation and stoichiometry. For aluminum carbonate (Al2(CO3)3), we identify aluminum (Al), carbon (C), and oxygen (O) as the constituent elements. Using a periodic table, we find the atomic masses of these elements are: Al (26.98 g/mol), C (12.01 g/mol), and O (16.00 g/mol).

To determine molar mass, we multiply each atomic mass by the number of times the element appears in the compound and add them up:\[ \text{Molar mass} = 2(26.98 \, \text{g/mol}) + 3(12.01 \, \text{g/mol}) + 9(16.00 \, \text{g/mol}) \]

This calculation yields the molar mass of aluminum carbonate, which is then used in subsequent calculations to convert from mass to moles or vice versa.

Stoichiometry

Stoichiometry sits at the heart of chemical reactions, as it involves the quantitative relationships between reactants and products. In this exercise, we particularly focus on the stoichiometry of aluminum carbonate dissociating to provide carbonate ions. To use stoichiometry effectively, we must understand the mole ratio dictated by the balanced chemical equation.

In the case of aluminum carbonate, each formula unit dissociates to release carbonate ions according to the following equation: \[ \text{Al}_2(\text{CO}_3)_3 \rightarrow 3\text{CO}_3^{2-} \]

From this, we infer a mole ratio of aluminum carbonate to carbonate ions as 1 to 3. This ratio allows us to calculate the amount of reactant or product formed in a chemical reaction. We apply this knowledge to determine the necessary volumes required in different situations, such as providing a specific amount of carbonate ions as needed in step 4 of the solution.

Solution Preparation

Preparing a chemical solution of a desired concentration requires careful measurement and calculation. Molarity (M), which is the number of moles of solute per liter of solution, serves as a common way to express solution concentration. In this task, we are instructed to prepare a 0.4500 M solution of aluminum carbonate.

To do this, we calculate the molarity by dividing the moles of solute (aluminum carbonate) by the volume of the solution in liters:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

By rearranging this equation, we can find the maximum volume of the solution that can be made from a given mass of solute. Adequate precision and attention to detail are crucial during this process to ensure the accuracy of the solution's concentration.

Mole Concept

The mole concept is a fundamental cornerstone in chemistry that relates the microscopic world of atoms and molecules to macroscopic quantities we can measure. One mole is Avogadro's number (approximately 6.022 x 10^23 particles) of any substance.

In practical terms, the mole allows chemists to count particles by weighing. For instance, we know that one mole of aluminum carbonate has a mass equal to its molar mass (calculated in the first section). This bridging between moles and mass is vital when it comes to finding out how much of a substance is needed or produced in a reaction.

Understanding the mole concept is particularly useful when we consider the ratio of different substances in a chemical reaction, as was needed in step 4 of the solution where we calculated the amount of solution necessary to provide a particular number of moles of carbonate ion.