stoichiometry pt1

Stoichiometry and Chemical Reactions

Overview

• Given: 65.3 grams of $N_2H_4$ (hydrazine).

• Goal: Find grams of $NH_3$ (ammonia).

• Important steps to convert from grams to moles and vice versa through stoichiometry.

Steps to Solve the Problem

1. Convert grams to moles of $N_2H_4$.

   • Molar Mass Calculation:

     • $N_2H_4$:

     • Nitrogen: $14 imes 2 = 28$

     • Hydrogen: $1 imes 4 = 4$

     • Total Molar Mass = $28 + 4 = 32$ grams/mole.

2. Stoichiometry Calculation:

   • Convert moles of $N_2H_4$ to moles of $NH_3$ using mole ratio coefficients from the balanced equation.

     • Coefficients: $N_2H_4$ (2) and $NH_3$ (4).

   • Use the relation:

     • $ext{Moles of } NH_3 = ext{Moles of } N_2H_4 imes rac{4}{2}$.

3. Convert moles of $NH_3$ to grams:

   • Molar Mass of $NH_3$:

     • Nitrogen: $14$

     • Hydrogen: $1 imes 3 = 3$

     • Total Molar Mass = $14 + 3 = 17$ grams/mole.

   • Use the calculated moles from step 2 to determine grams:

     • $ext{Grams of } NH_3 = ext{Moles of } NH_3 imes 17 ext{ grams/mole}$.

Theoretical Yield

• Definition:

  • The maximum yield expected from a chemical reaction under ideal conditions, calculated using stoichiometry.

• Experimental Yield:

  • The amount of product actually obtained from a reaction performed in the laboratory; usually less than theoretical yield due to errors.

• Percent Yield Calculation:

  • Formula:
    $ext{Percent Yield} = rac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100$.

  • Example yields: 90%, 50%, 20% indicate reaction efficiency.

• Key Points:

  • Theoretical yield is calculated, while actual yield is provided.

  • Actual yield can vary due to experimental error.

Limiting Reactant Problems

• Definition:

  • The reactant that limits the amount of product formed in a chemical reaction; it will produce the smallest amount of product.

• Problem Example:

  • Given: 4 moles of aluminum and 2.6 moles of oxygen for the reaction to form $Al_2O_3$.

• Procedure:

  1. Determine how much $Al_2O_3$ can be made from each reactant (aluminum and oxygen).

  • Use stoichiometry coefficients from the balanced equation (
    $4Al + 3O_2 <br>ightarrow 2Al_2O_3$).

  1. Apply mole conversions for each possible reactant:

  • Starting with Aluminum:

    • $ext{Moles of } Al_2O_3 = 4 ext{ moles of } Al imes rac{2 ext{ moles of } Al_2O_3}{4 ext{ moles of } Al} = 2 ext{ moles of } Al_2O_3$.

  • Starting with Oxygen:

    • $ext{Moles of } Al_2O_3 = 2.6 ext{ moles of } O_2 imes rac{2}{3} = 1.73 ext{ moles of } Al_2O_3$.

    • Compare results:

    • Smaller amount indicates the limiting reactant:

    • 1.73 moles from Oxygen means oxygen is the limiting reactant.

Concepts on Solutions

Definitions

• Solution:

  • A homogeneous mixture formed when a solute is dissolved in a solvent.

• Solvent:

  • The major component (e.g., water) in which the solute is dissolved.

• Solute:

  • The substance (e.g., sodium chloride) that is dissolved in the solvent.

Concentration

• Importance of Concentration:

  • Indicates the amount of solute present in a given volume of solution.

  • Concentrations can be classified as:

  • High Concentration (Concentrated)

  • Low Concentration (Dilute)

• Concentration can be quantified using Molarity (M):

  • Defined as moles of solute per liter of solution.

  • Formula:
    $ext{Molarity (M)} = rac{ ext{moles of solute}}{ ext{liters of solution}}$.

Practical Application Examples

1. Calculating Molarity:

   • Given moles and volume:

     • If you have 0.45 moles of solute in 1.5 liters of solution,

     • $M = rac{0.45}{1.5} = 0.30 ext{ M}$.

2. Conversions for Calculating Molarity:

   • Convert grams to moles and milliliters to liters when necessary for calculations.

Conclusion

• It's essential to understand stoichiometry calculations, reactions, theoretical and experimental yield, limiting Reactants, and the concept of solutions with regard to concentration and molarity.