150 mL of 0.583 M potassium chloride (74.55 g/mol) reacts with an excess of lead (II) nitrate (325.39 g/mol). Write the net ionic equation for this reaction. Clearly indicate charges and states of matter as appropriate. b: How many molecules of precipitate are theoretically formed in this reaction?
The Correct Answer and Explanation is:
Let’s break this down step by step.
Part A: Writing the Net Ionic Equation
1. Write the balanced molecular equation:
The reaction between potassium chloride (KCl) and lead (II) nitrate (Pb(NO₃)₂) forms lead (II) chloride (PbCl₂) as a precipitate and potassium nitrate (KNO₃) in solution:
Pb(NO₃)₂ (aq)+2KCl (aq)→PbCl₂ (s)+2KNO₃ (aq)\text{Pb(NO₃)₂ (aq)} + 2\text{KCl (aq)} \rightarrow \text{PbCl₂ (s)} + 2\text{KNO₃ (aq)}Pb(NO₃)₂ (aq)+2KCl (aq)→PbCl₂ (s)+2KNO₃ (aq)
2. Dissociate the aqueous compounds into ions:
- Pb(NO₃)₂ (aq) dissociates into Pb²⁺ and 2NO₃⁻
- KCl (aq) dissociates into K⁺ and Cl⁻
- KNO₃ (aq) dissociates into K⁺ and NO₃⁻
The net ionic equation will involve only the ions that participate in the reaction.
3. Eliminate spectator ions:
Spectator ions are ions that do not change during the reaction. In this case, the potassium ion (K⁺) and the nitrate ion (NO₃⁻) are spectator ions. Therefore, they are removed from the net ionic equation.
Net ionic equation:Pb²⁺ (aq)+2Cl⁻ (aq)→PbCl₂ (s)\text{Pb²⁺ (aq)} + 2\text{Cl⁻ (aq)} \rightarrow \text{PbCl₂ (s)}Pb²⁺ (aq)+2Cl⁻ (aq)→PbCl₂ (s)
This shows the formation of lead (II) chloride as a precipitate.
Part B: Calculating the Number of Molecules of Precipitate
1. Moles of potassium chloride (KCl):
We are given 150 mL of 0.583 M KCl. First, calculate the moles of KCl:moles of KCl=0.150 L×0.583 mol/L=0.08745 mol of KCl\text{moles of KCl} = 0.150 \, \text{L} \times 0.583 \, \text{mol/L} = 0.08745 \, \text{mol of KCl}moles of KCl=0.150L×0.583mol/L=0.08745mol of KCl
Since the stoichiometric ratio between KCl and PbCl₂ is 2:1, the moles of PbCl₂ formed will be half of the moles of KCl:moles of PbCl₂=0.08745 mol of KCl2=0.043725 mol of PbCl₂\text{moles of PbCl₂} = \frac{0.08745 \, \text{mol of KCl}}{2} = 0.043725 \, \text{mol of PbCl₂}moles of PbCl₂=20.08745mol of KCl=0.043725mol of PbCl₂
2. Calculate the number of molecules of PbCl₂:
Now, use Avogadro’s number (6.022 × 10²³ molecules/mol) to find the number of molecules of PbCl₂:molecules of PbCl₂=0.043725 mol of PbCl₂×6.022×1023 molecules/mol\text{molecules of PbCl₂} = 0.043725 \, \text{mol of PbCl₂} \times 6.022 \times 10^{23} \, \text{molecules/mol}molecules of PbCl₂=0.043725mol of PbCl₂×6.022×1023molecules/molmolecules of PbCl₂=2.63×1022 molecules of PbCl₂\text{molecules of PbCl₂} = 2.63 \times 10^{22} \, \text{molecules of PbCl₂}molecules of PbCl₂=2.63×1022molecules of PbCl₂
Summary:
- The net ionic equation for the reaction is:
Pb²⁺ (aq)+2Cl⁻ (aq)→PbCl₂ (s)\text{Pb²⁺ (aq)} + 2\text{Cl⁻ (aq)} \rightarrow \text{PbCl₂ (s)}Pb²⁺ (aq)+2Cl⁻ (aq)→PbCl₂ (s) - The number of molecules of PbCl₂ theoretically formed is:
2.63×1022 molecules2.63 \times 10^{22} \, \text{molecules}2.63×1022molecules
