Problem 105 - Chemistry: The Central Science in SI Units, Global Edition

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

In summary: a) Zinc reacts with the copper(II) nitrate solution. b) The net ionic equation is: \(Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}\) c) Zinc is the limiting reagent. d) The molarity of the remaining Cu²⁺ ions in the solution is 0.108 M.

Step 1: Determine which metal will react with the copper(II) nitrate solution

To determine which metal will react with Cu(NO₃)₂, we need to consult the activity series of metals. The activity series is a list that ranks metals based on their tendency to undergo redox reactions. A metal higher in the series will displace a metal ion lower in the series from its salt in an aqueous solution. The activity series for some common metals is: Mg > Al > Zn > Cr > Fe > Co > Ni > Sn > Pb > H > Cu > Ag Zinc (Zn) is higher in the activity series than copper (Cu), while silver (Ag) is lower. Therefore, zinc will react with the copper(II) nitrate solution, while silver will not.

Step 2: Write the balanced net ionic equation

The reaction between zinc and copper(II) nitrate is a single displacement reaction, in which, zinc displaces copper from its nitrate salt as follows: \(Zn(s) + Cu(NO_3)_2(aq) \rightarrow Zn(NO_3)_2(aq) + Cu(s)\) Now, let's write the net ionic equation for the reaction. First, separate all the soluble compounds into ions: \([Zn(s) \rightarrow Zn_{(s)}]\) \([Cu(NO_3)_2(aq) \rightarrow Cu^{2+}_{(aq)}+2NO_3^{-}_{(aq)}]\) \([Zn(NO_3)_2(aq) \rightarrow Zn^{2+}_{(aq)}+2NO_3^{-}_{(aq)}]\) Since nitrate ions are present on both sides, they are spectator ions and can be eliminated from the equation: Net ionic equation: \(Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}\)

Step 3: Identify the limiting reagent

We are given 3.00 g of zinc and a 500.0 mL solution of 0.2 M copper(II) nitrate. To determine the limiting reagent, we first need to calculate the moles of each reactant. Moles of zinc: \(\frac{3.00 \ \mathrm{g}}{65.38 \ \mathrm{g\ mol^{-1}}}=0.0459 \ \mathrm{mol}\) (using the molar mass of zinc) Moles of copper(II) nitrate: \(\mathrm{0.2 \ M} \times \mathrm{0.500 \ L} = 0.100 \ \mathrm{mol}\) Now, compare the stoichiometry of the balanced reaction: \(1 \ \mathrm{mol} \ Zn : 1 \ \mathrm{mol} \ Cu(NO_3)_2\) \(0.0459 \ \mathrm{mol} \ Zn : 0.100 \ \mathrm{mol} \ Cu(NO_3)_2\) Since there is less zinc than copper(II) nitrate in terms of their stoichiometry, zinc is the limiting reagent.

Step 4: Calculate the molarity of Cu²⁺ ions in the resulting solution

To calculate the molarity of the remaining Cu²⁺ ions in the solution, we first need to find the moles of Cu²⁺ ions that did not participate in the reaction. Moles of reacted Cu²⁺ ions: \(0.0459 \ \mathrm{mol}\) (based on the moles of limiting reagent – zinc) Moles of unreacted Cu²⁺ ions: \(0.100 \ \mathrm{mol} - 0.0459 \ \mathrm{mol} = 0.0541 \ \mathrm{mol}\) Now, we can calculate the molarity of Cu²⁺ ions in the resulting solution: Molarity of Cu²⁺ ions: \(\frac{0.0541 \mathrm{\ mol}}{0.500 \mathrm{\ L}} = 0.108 \mathrm{M}\) In conclusion: a) Zinc will react with the copper(II) nitrate solution. b) The net ionic equation for the reaction is: \(Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}\) c) The limiting reagent is zinc. d) The molarity of Cu²⁺ ions in the resulting solution is 0.108 M.

Key Concepts

Net Ionic Equation

In a chemical reaction, especially in an aqueous solution, we often focus on the substances that are actively participating in the observed chemical change. Here enters the concept of the net ionic equation, which simplifies the equations of reactions by stripping away the spectator ions—those ions that do not participate in the actual redox process.
For instance, let's consider the reaction between zinc metal and copper(II) nitrate solution. Initially, the full equation includes not only the elements of zinc and copper but also the nitrate ions:

• Full equation: \(Zn(s) + Cu(NO_3)_2(aq) \rightarrow Zn(NO_3)_2(aq) + Cu(s)\)

After identifying the spectator ions, which are the nitrate ions \(NO_3^−\) in this case, we simplify the equation to include only the species that participate in the reaction:

• Net ionic equation: \(Zn(s) + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu(s)\)

Doing this allows for better clarity regarding which elements and ions are undergoing the chemical change, making understanding and solving questions simpler.

Limiting Reagent

The concept of a limiting reagent is crucial in determining the extent of a chemical reaction. Simply put, the limiting reagent is the substance that is completely consumed first, thus determining when the reaction stops and how much product is formed.
To identify the limiting reagent, you need to analyze the stoichiometry of the reaction. In the example involving zinc and copper(II) nitrate, you start by calculating the moles of each reactant:

• Moles of zinc: \(\frac{3.00 \ \text{g}}{65.38 \ \text{g/mol}} = 0.0459 \ \text{mol}\)
• Moles of copper(II) nitrate: \(0.2 \ \text{M} \times 0.500 \ \text{L} = 0.100 \ \text{mol}\)

The balanced equation shows a 1:1 ratio for zinc and copper(II) nitrate reaction. Because zinc is present in a smaller molar amount compared to copper(II) nitrate, zinc acts as the limiting reagent. Once the zinc is fully consumed, no further reaction will occur, highlighting its vital role in determining the reaction's outcome.

Molarity Calculation

Molarity, defined as moles of solute per liter of solution, is a key concept used to describe the concentration of a solution. This can be particularly useful when determining the concentration of ions left in a solution after a reaction has occurred.
For the reaction between zinc and copper(II) nitrate, once the reaction is complete, determining the molarity of copper(II) ions that remain gives insight into how the reaction impacts solution concentration.
The initial moles of Cu²⁺ ions are given by the initial concentration and volumes: \(0.2 \ \text{M} \times 0.500 \ \text{L} = 0.100 \ \text{mol}\). After the reaction with zinc, 0.0459 mol of copper ions have reacted.
Moles of unreacted Cu²⁺ ions is calculated as follows:

• Unreacted moles: \(0.100 \ \text{mol} - 0.0459 \ \text{mol} = 0.0541 \ \text{mol}\)

Thus, the molarity of Cu²⁺ ions remaining in the solution is \(\frac{0.0541 \ \text{mol}}{0.500 \ \text{L}} = 0.108 \ \text{M}\), providing the concentration of copper ions left after the chemical reaction.