Problem 30 - Chemistry Matter and Change

Question

Challenge When copper wire is placed into a silver nitrate solution \(\left(\mathrm{AgNO}_{3}\right),\) silver crystals and copper(ll) nitrate \(\left(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\right)\) solution form. \begin{equation} \begin{array}{l}{\text { a. Write the balanced chemical equation for the reaction. }} \\ {\text { b. If a } 20.0 \text { -g sample of copper is used, determine the theoretical yield }} \\ {\text { of silver. }} \\ {\text { c. If } 60.0 \text { g of silver is recovered from the reaction, determine the }} \\\ {\text { percent yield of the reaction. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

Balanced equation: \( \text{Cu} + 2\text{AgNO}_3 \rightarrow \text{Cu(NO}_3\text{)}_2 + 2\text{Ag} \). Theoretical yield: 67.93 g of Ag. Percent yield: 88.3\%.

Step 1: Write the Balanced Chemical Equation

The reaction between copper wire and silver nitrate is a single replacement reaction. The copper displaces silver from silver nitrate to form copper(II) nitrate and silver. The unbalanced equation is: \[ \text{Cu (s)} + \text{AgNO}_3 (aq) \rightarrow \text{Cu(NO}_3\text{)}_2 (aq) + \text{Ag (s)} \] To balance this equation, note that one copper replaces two silver ions, yielding the balanced equation: \[ \text{Cu (s)} + 2\text{AgNO}_3 (aq) \rightarrow \text{Cu(NO}_3\text{)}_2 (aq) + 2\text{Ag (s)} \]

Step 2: Calculate the Theoretical Yield of Silver

First, calculate the moles of copper used: \( 20.0 \text{ g Cu} \times \frac{1 \text{ mole Cu}}{63.55 \text{ g/mol}} \approx 0.3148 \text{ moles Cu} \). From the balanced equation, 1 mole of Cu produces 2 moles of Ag. Hence, \( 0.3148 \text{ moles Cu} \times \frac{2 \text{ moles Ag}}{1 \text{ mole Cu}} = 0.6296 \text{ moles Ag} \). Convert moles of silver to grams using the molar mass of Ag: \( 0.6296 \text{ moles Ag} \times 107.87 \text{ g/mol} \approx 67.93 \text{ g Ag} \). This is the theoretical yield.

Step 3: Determine the Percent Yield

Percent yield is calculated using the formula: \( \text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% \). The actual yield is given as 60.0 g of silver. So, \( \text{Percent Yield} = \left( \frac{60.0}{67.93} \right) \times 100\% \approx 88.3\% \).

Key Concepts

Single Replacement Reaction

In a single replacement reaction, one element replaces another element in a compound, and this type of chemical reaction is also known as a single displacement reaction. These reactions are characterized by the element swapping places with one part of a compound.
For example, in the reaction of copper wire with silver nitrate, copper takes the place of silver, leading to the formation of copper(II) nitrate and silver metal. The general form of a single replacement reaction can be expressed as:

• \[ A + BC ightarrow AC + B \]

In this equation, element \( A \) replaces \( B \) in the compound \( BC \). This process often involves metals and nonmetals, where a more reactive metal displaces a less reactive one.
To predict whether a single replacement will occur, the reactivity series can be used. In our case, copper is more reactive than silver, making the reaction possible. Understanding this concept is fundamental to predicting the outcome of similar reactions.

Balanced Chemical Equation

Balancing chemical equations is crucial for representing chemical reactions accurately. It ensures that the same number of each type of atom appears on both sides of the equation, which reflects the law of conservation of mass.
For our reaction involving copper and silver nitrate, the unbalanced equation initially looks like this:

• \[ \text{Cu (s)} + \text{AgNO}_3 (aq) \rightarrow \text{Cu(NO}_3\text{)}_2 (aq) + \text{Ag (s)} \]

In order to balance it, you need to have the same number of silver atoms in the reactants and the products. Copper replaces two silver ions, hence the balanced version of this equation becomes:

• \[ \text{Cu (s)} + 2\text{AgNO}_3 (aq) \rightarrow \text{Cu(NO}_3\text{)}_2 (aq) + 2\text{Ag (s)} \]

The key to balancing equations is to count the atoms of each element and adjust coefficients as required, making sure the atoms are balanced on both sides. This balanced equation not only shows how many of each molecule participates but also forms the basis for determining stoichiometric relationships and yield calculations.

Theoretical Yield

The theoretical yield of a chemical reaction is the maximum amount of product that can be produced from a given amount of reactant, based on stoichiometric calculations. It assumes that the reaction occurs perfectly and completely, without any losses or side reactions.
To determine the theoretical yield for the reaction between copper and silver nitrate, we first calculate the moles of copper:

• Given 20.0 g of Cu and using the molar mass (63.55 g/mol), the moles of Cu are calculated as: \[ 20.0 \text{ g Cu} \times \frac{1 \text{ mole Cu}}{63.55 \text{ g/mol}} \approx 0.3148 \text{ moles Cu} \]

Based on the balanced equation, 1 mole of copper yields 2 moles of silver. Thus, converting is as follows:

• \[ 0.3148 \text{ moles Cu} \times \frac{2 \text{ moles Ag}}{1 \text{ mole Cu}} = 0.6296 \text{ moles Ag} \]

Converting these moles to grams using the molar mass of silver (107.87 g/mol), gives us:

• \[ 0.6296 \text{ moles Ag} \times 107.87 \text{ g/mol} \approx 67.93 \text{ g Ag} \]

This is the theoretical yield, indicating under perfect conditions, you'd expect to obtain 67.93 g of silver from the reaction.

Percent Yield

Percent yield is an important aspect of quantifying the efficiency of a chemical reaction. It compares the actual yield, the amount of product actually obtained from the reaction, to the theoretical yield you calculated.
This is expressed as a percentage and is calculated using the formula:

• \[ \text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% \]

In our scenario, the actual yield is given as 60.0 g of silver, while the theoretical yield, as calculated, is 67.93 g. Using the formula, the percent yield is:

• \[ \text{Percent Yield} = \left( \frac{60.0}{67.93} \right) \times 100\% \approx 88.3\% \]

A percent yield less than 100% typically suggests inefficiencies, such as incomplete reactions or product losses. Conversely, a yield over 100% could indicate measurement errors or impurities. Understanding percent yield helps in optimizing reaction conditions and evaluating reaction efficiency in a laboratory or industrial setting.