Stoichiometry Lecture Notes

Introduction to Stoichiometry

• Revisiting stoichiometry after covering concentration and solutions.
• Central concept: Moles with a focus on mole ratio.
• Different pathways to reach moles:
  • Mass
  • Volume of gas at standard temperature and pressure (STP)
  • Representative particles
• Molarity: Molarity = Moles / Liters
  • Use to find moles of solute from concentration and volume.

Example Problem: Ammonium Nitrate Production

• Question: How many grams of ammonium nitrate can be produced from 55.0 mL of 3.2 M ammonium carbonate solution reacting with 100 mL of excess aqueous copper (II) nitrate?
• Steps to Solve:
  1. Balanced Chemical Equation: Write down the balanced equation.
  2. Identify Known and Unknown:
     • Known: 55 mL of 3.2 M solution.
     • Unknown: Grams of ammonium nitrate.
  3. Determine Molar Mass: Calculate molar mass of ammonium nitrate (80.06 g/mol).
  4. Conversion Steps:
     • Use volume and concentration to find moles of ammonium carbonate.
     • Apply mole ratio from balanced equation (1:2 ratio of ammonium carbonate to ammonium nitrate).
     • Convert from moles to grams using molar mass.
  5. Calculation:
     • Volume of solution used: 55 mL (0.055 L).
     • Moles of ammonium carbonate: 0.055 L * 3.2 mol/L = 0.176 mol.
     • Moles of ammonium nitrate produced: 0.176 mol * 2 = 0.352 mol.
     • Grams of ammonium nitrate: 0.352 mol * 80.06 g/mol = 28.15 g.*

Final Concentration Calculation

• Question: What is the concentration of ammonium nitrate in the final solution?
• Approach:
  • Use calculated grams of ammonium nitrate to find moles.
  • Total volume of solution: 55 mL + 100 mL = 155 mL (0.155 L).
  • Moles of ammonium nitrate: 0.352 mol.
  • Concentration: Molarity = Moles / Volume = 0.352 mol / 0.155 L = 2.27 M.

Limiting Reactant Problem Example

• Question: Identify the limiting reactant in a reaction between silver nitrate and magnesium chloride.
• Balanced Chemical Equation: 2 AgNO₃ + MgCl₂ → 2 AgCl + Mg(NO₃)₂
• Given Information:
  • Concentrations of reactants provided, to convert to moles:
    • AgNO₃: 0.150 moles
    • MgCl₂: 0.087875 moles
• Steps:
  1. Start with moles of one reactant (e.g., AgNO₃).
  2. Use mole ratio (1:1) to find equivalent moles of the second reactant (MgCl₂).
  3. Compare available moles with required moles to identify limiting reactant.
• Result: AgNO₃ is the limiting reagent and MgCl₂ is in excess.

Conclusion

• Reviewed stoichiometric calculations involving solutions.
• Practiced identifying limiting reactants in reactions.
• Gained new skills in using concentration and volume in stoichiometry.