Stoichiometry and Solutions

Overview of Stoichiometry

• Center of stoichiometry map is moles with mole ratio at the center.
• Various methods to reach moles:
  • Mass
  • Volume of gas (at STP)
  • Representative particles
• Molarity: Moles per liter (M = moles/L)
  • Knowing concentration & volume allows calculation of moles of solute.

Example Problem: Ammonium Nitrate Production

Given Information

• Volume of ammonium carbonate solution: 55.0 mL
• Concentration of ammonium carbonate: 3.2 M
• Volume of excess copper nitrate: 100 mL

Steps to Solve

1. Balanced Chemical Equation: Required for all stoichiometric problems.

   • Ammonium carbonate + copper nitrate → Ammonium nitrate + copper carbonate (precipitate)

2. Identify Known and Unknown:

   • Known: Volume (55 mL), Concentration (3.2 M)
   • Unknown: Grams of ammonium nitrate produced.

3. Calculate Moles of Reactant:

   • Convert volume to liters: 55 mL = 0.055 L
   • Calculate moles of ammonium carbonate:
     • Moles = Volume (L) × Concentration (M) = 0.055 L × 3.2 M = 0.176 moles

4. Mole Ratio:

   • From equation: 1 mole ammonium carbonate to 1 mole ammonium nitrate.
   • Moles of ammonium nitrate = 0.176 moles

5. Convert Moles to Grams:

   • Molar mass of ammonium nitrate = 80.06 g/mol
   • Grams = Moles × Molar Mass = 0.176 moles × 80.06 g/mol = 14.1 g

6. Final Calculation: Should arrive at 28 grams of ammonium nitrate.

Bonus Question

• Calculate concentration of ammonium nitrate in the final solution:
  • Total volume = 55 mL + 100 mL = 155 mL = 0.155 L
  • Moles of ammonium nitrate = 0.176 moles
  • Concentration = Moles / Volume = 0.176 moles / 0.155 L = 1.135 M

Identifying Limiting Reactant Problem

Given Information

• Balanced chemical equation:
  • 2 moles AgNO3 + 1 mole MgCl2 → 2 moles AgCl + 1 mole Mg(NO3)2
• Concentrations:
  • AgNO3: 0.150 M
  • MgCl2: 0.875 M

Steps to Solve

1. Convert Concentrations to Moles:

   • AgNO3: 0.150 M × Volume (assumed 1 L for simplicity) = 0.150 moles
   • MgCl2: 0.875 M × Volume (assumed 1 L for simplicity) = 0.875 moles

2. Mole Ratio for Limiting Reagent:

   • Start with AgNO3:
     • (0.150 moles AgNO3) × (1 mole MgCl2 / 2 moles AgNO3) = 0.075 moles MgCl2 needed
   • Since you have 0.875 moles of MgCl2, AgNO3 is the limiting reagent.

Conclusion

• Final identification: Silver nitrate is the limiting reagent.
• Overview of using solutions in stoichiometric problems.

Key Concepts

• Understand the stoichiometry map for effective calculations.
• Recognize the importance of balanced equations and mole ratios in problem-solving.
• Practice with different problems to solidify understanding.