Problem 105 - Chemistry The Central Science In Si Units

Question

Suppose you have \(3.00 \mathrm{~g}\) of powdered zinc metal, \(3.00 \mathrm{~g}\) of powdered silver metal and \(500.0 \mathrm{~mL}\) of a \(0.2 \mathrm{M}\) copper(II) nitrate solution. (a) Which metal will react with the copper(II) nitrate solution? (b) What is the net ionic equation that describes this reaction? (c) Which is the limiting reagent in the reaction? (d) What is the molarity of \(\mathrm{Cu}^{2+}\) ions in the resulting solution?

Step-by-Step Solution

Answer

(a) Zinc will react. (b) \( \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \). (c) Zinc is the limiting reagent. (d) Molarity of \( \text{Cu}^{2+} \) is 0.1082 M.

Step 1: Identify Metals Reactivity

To determine which metal will react, consider the reactivity series. Zinc is more reactive than copper whereas silver is less reactive. Thus, zinc will react with copper(II) nitrate, but silver will not.

Step 2: Write the Net Ionic Equation

Since zinc will react, we need to write the reaction of zinc with copper(II) nitrate. The balanced net ionic equation is:\[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \]This equation shows that zinc displaces copper from its solution.

Step 3: Calculate Amount of Reactants

Calculate the moles of zinc available using its atomic mass (65.38 g/mol):\[ \text{Moles of Zn} = \frac{3.00 \text{ g}}{65.38 \text{ g/mol}} = 0.0459 \text{ mol} \]Calculate the moles of copper ions using molarity and volume:\[ \text{Moles of Cu}^{2+} = 0.2 \text{ M} \times 0.500 \text{ L} = 0.1 \text{ mol} \]

Step 4: Determine the Limiting Reagent

To determine the limiting reagent, compare the mole ratio from the balanced equation. The reaction requires a 1:1 ratio of Zn to Cu²⁺. Zinc provides 0.0459 mol, whereas copper ions are 0.1 mol. Therefore, zinc is the limiting reagent.

Step 5: Calculate Remaining Concentration of Cu²⁺

Since zinc is the limiting reagent, all 0.0459 mol of Zn will react, consuming an equal amount of Cu²⁺, leaving:\[ 0.1 \text{ mol} - 0.0459 \text{ mol} = 0.0541 \text{ mol of Cu}^{2+} \]Calculate the remaining molarity of Cu²⁺ ions in 500 mL:\[ \text{Molarity of Cu}^{2+} = \frac{0.0541 \text{ mol}}{0.500 \text{ L}} = 0.1082 \text{ M} \]

Key Concepts

Reactivity Series

Understanding the reactivity series is crucial in predicting which metals will react in a chemical reaction. The reactivity series is a list of metals arranged in order of their reactivity from highest to lowest. More reactive metals tend to lose electrons and form positive ions more readily than less reactive metals. In our exercise, zinc is more reactive than copper, which means it can displace copper ions from copper(II) nitrate. Silver, being less reactive than both zinc and copper, will not react in this scenario. Essentially, knowing this series allows us to identify that zinc will engage in a chemical reaction, displacing copper ions and forming copper metal in the process. When assessing chemical reactions involving metals, always refer back to the reactivity series to determine potential reactions.

Net Ionic Equations

Net ionic equations simplify chemical reactions by showing only the particles involved in the actual chemical change. This type of equation eliminates spectator ions, or ions that remain unchanged during the reaction, from the complete ionic equation. For the zinc and copper(II) nitrate reaction:

• Begin by writing the balanced molecular equation.
• Identify spectator ions, eliminating them to focus on the ions and compounds that undergo change.
• The resulting net ionic equation for zinc reacting with copper(II) nitrate is: \[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \]

This equation effectively highlights that zinc is oxidized to Zn²⁺, while Cu²⁺ is reduced to copper metal. Writing net ionic equations helps to clearly visualize the chemical process and identify changes to each species involved.

Limiting Reagents

In a chemical reaction, the limiting reagent is the reactant that is entirely consumed first, limiting the amount of product formed. To find the limiting reagent, compare the mole ratio of the reactants based on the balanced chemical equation. In our example:

• Calculate the moles of each reactant available: zinc is 0.0459 mol, and Cu²⁺ is 0.1 mol.
• Determine the stoichiometry of the reaction: the reaction requires a 1:1 ratio of zinc to copper ions.
• Since zinc provides fewer moles than Cu²⁺, it is the limiting reagent.

Identifying the limiting reagent is important as it dictates the maximum yield of product. In this case, the zinc limits the formation of copper metal, which impacts calculations for the remaining concentration of copper ions after the reaction.

Molarity Calculations

Molarity is a measure of the concentration of a solute in a solution. It is described as moles of solute per liter of solution, denoted as M. In the exercise, it's important for calculating how much of the reactant is used, and how much of the product and remaining reactant are in the solution. In our example:

• After the reaction, you have unreacted Cu²⁺ left in the solution.
• To find the molarity of Cu²⁺ remaining, subtract moles of zinc reacted from initial moles of Cu²⁺, which is 0.1 mol - 0.0459 mol, leaving 0.0541 mol of Cu²⁺.
• Divide by the solution volume, 0.5 L, to find the new molarity of Cu²⁺:
\[ \text{Molarity of Cu}^{2+} = \frac{0.0541 \, \text{mol}}{0.500 \, \text{L}} = 0.1082 \, \text{M} \]

This calculation allows you to understand the remaining concentration of ions in a solution after the reaction, which is crucial for both practical and theoretical chemistry applications.