Problem 50 - General Chemistry: Principles and Modern Applications

Question

Excess \(\mathrm{NaHCO}_{3}\) is added to \(525 \mathrm{mL}\) of \(0.220 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} .\) These substances react as follows: \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow\) $$ \mathrm{CuCO}_{3}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ (a) How many grams of the \(\mathrm{NaHCO}_{3}(\mathrm{s})\) will be consumed? (b) How many grams of \(\mathrm{CuCO}_{3}(\mathrm{s})\) will be produced?

Step-by-Step Solution

Answer

(a) 19.41 g of NaHCO3 will be consumed. (b) 14.26 g of CuCO3 will be produced.

Step 1: Determine moles of copper nitrate [Cu(NO3)2]

We know the volume (0.525 L) and molarity (0.220 M) of the copper nitrate solution. Apply the equation ‘n=c×V’ where 'n' is the number of moles, 'c' is the molarity and 'V' is the volume in liter. Calculate as 'n = 0.220 M × 0.525 L = 0.1155 mol'. So, we have 0.1155 mol of copper nitrate.

Step 2: Determine moles of sodium bicarbonate (NaHCO3) that will be consumed

From the balanced chemical equation, we see that 1 mol of Cu(NO3)2 reacts with 2 mol of NaHCO3. Therefore, the amount of NaHCO3 consumed will be twice the amount of Cu(NO3)2. Calculate as 'n(NaHCO3) = 2 * n(Cu(NO3)2) = 2 * 0.1155 mol = 0.231 mol'

Step 3: Compute the mass of NaHCO3 consumed

Convert the moles of NaHCO3 to grams using its molar mass (84 g/mol). Use the equation ‘m = n × M’ where 'm' is the mass, 'n' is the number of moles and 'M' is the molar mass. Calculate as 'm(NaHCO3) = 0.231 mol * 84 g/mol = 19.41 g'

Step 4: Determine moles of copper carbonate (CuCO3) that will be produced

From the balanced chemical equation, 1 mol of Cu(NO3)2 produces 1 mol CuCO3. So the amount of CuCO3 produced will be the same as the amount of Cu(NO3)2. Therefore, n(CuCO3) = n(Cu(NO3)2) = 0.1155 mol.

Step 5: Compute the mass of CuCO3 produced

Convert the moles of CuCO3 to grams using its molar mass (123.5 g/mol). Calculate as 'm(CuCO3) = n(CuCO3) * M(CuCO3) = 0.1155 mol * 123.5 g/mol = 14.26 g'

Key Concepts

Molarity

Molarity is a crucial concept in chemistry that helps us understand the concentration of a solution. It is defined as the number of moles of solute per liter of solution. In our exercise, we use molarity to calculate how much copper nitrate is in the solution. Imagine mixing a certain amount of copper nitrate in some water until you reach 525 mL. Here, the solution has a molarity of 0.220 M, meaning there are 0.220 moles of copper nitrate in every liter.
To find the exact number of moles in our 525 mL (or 0.525 L), we use the formula: \[ n = c \times V \]where:

• \(n\) is the number of moles
• \(c\) is the molarity
• \(V\) is the volume in liters

Using this equation, we find there are 0.1155 moles of copper nitrate in the solution.
Understanding molarity is essential for determining how substances will react in a given situation. Always measure volumes in liters when working with molarity to maintain consistency.

Balanced Chemical Equation

A balanced chemical equation is like a recipe for a chemical reaction. It shows us how reactants transform into products. In this exercise, the equation tells us how copper nitrate reacts with sodium bicarbonate to form copper carbonate and other products. The balanced equation is:\[\text{Cu(NO}_3\text{)}_2\text{(aq)} + 2\ \text{NaHCO}_3\text{(s)} \rightarrow \text{CuCO}_3\text{(s)} + 2\ \text{NaNO}_3\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\]This equation shows us the proportions in which reactants combine and products form.
For instance, one mole of copper nitrate reacts with two moles of sodium bicarbonate. Notice that the equation balances the number of atoms on each side, following the law of conservation of mass. This is crucial because atoms do not disappear or appear out of nowhere during a chemical reaction.
To use the balanced equation effectively, always ensure each compound's stoichiometric coefficients correctly reflect its role in the reaction. By referring to these coefficients, you can easily calculate the amounts of reactants and products involved. Balancing equations ensures that chemical reactions proceed with no leftover substances unaccounted for.

Mole Concept

The mole concept is foundational in chemistry. It allows chemists to count numbers of atoms, molecules, or compounds by weighing them. One mole of any substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\). It serves as a bridge between the microscopic world of atoms and the macroscopic world we experience.
In our problem, we use the mole concept to determine how much sodium bicarbonate will react and how much copper carbonate will be formed. For example, we calculated that 0.1155 moles of copper nitrate were used, leading us to deduce that 0.231 moles of sodium bicarbonate would be consumed, based on the balanced equation.
Using the concept further, we determined that 0.1155 moles of copper carbonate would be produced since the reaction shows a 1:1 molar ratio between copper nitrate and copper carbonate.

Converting Moles to Grams: The conversion process involves using the molar mass, which you can think of as the weight of one mole of a substance. Each substance has its unique molar mass in grams per mole, like NaHCO\(_3\) having 84 g/mol and CuCO\(_3\) having 123.5 g/mol. We multiply the moles of a substance by its molar mass to find the mass in grams:\[\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}\]This calculation helps us convert theoretical chemical equations into practical measurements in the lab. Understanding how to relate moles to grams using the molar mass is vital for quantifying chemical reactions.