9/23/25 Chemistry Lecture Chapter 4: Stoichiometry & Chapter 5: Solutions and Aqueous Reactions Study Notes

Chapter 4: Stoichiometry Review

Quiz and Homework Information

• Homework is due Wednesday at 4 PM, with unlimited redos after.
• A quiz covering Chapter 4 will be given at the beginning of Thursday's class.

Stoichiometry Problem Walkthrough

This section reviews a comprehensive stoichiometry problem, addressing theoretical yield, limiting reactants, percent yield, and excess reagents.

Problem Setup

• Given: $45.69 \text{ grams}$ of phosphorus gas reacts with $131.3 \text{ grams}$ of fluorine gas.
• Produced: $76.3 \text{ grams}$ of phosphorus trichloride gas is actually produced.
• Questions to answer:
  1. What is the theoretical yield?
  2. What is the limiting reactant?
  3. What is the percent yield?
  4. How much excess reagent remains?

1. Writing and Balancing the Chemical Equation

• Phosphorus gas: Not just P. It's a polyatomic molecule, $\text{P}_4$.
• Chlorine gas: A diatomic molecule, $\text{Cl}_2$.
• Phosphorus trichloride gas: $\text{PCl}_3$.
• Unbalanced Equation: $\text{P}<em>4(g) + \text{Cl}</em>2(g) \rightarrow \text{PCl}_3(g)$
• Balancing Steps:
  • Balance P: Four P atoms on the left, one on the right. Add a coefficient of 4 to $\text{PCl}<em>3$. $\text{P}</em>4(g) + \text{Cl}<em>2(g) \rightarrow 4\text{PCl}</em>3(g)$
  • Balance Cl: Now there are $4 \times 3 = 12$ Cl atoms on the right. Add a coefficient of 6 to $\text{Cl}<em>2$ on the left. $\text{P}</em>4(g) + 6\text{Cl}<em>2(g) \rightarrow 4\text{PCl}</em>3(g)$
• Balanced Chemical Equation: This equation provides the mole ratios for conversions.
  • $1 \text{ mole of P}<em>4$ (s) reacts with $6 \text{ moles of Cl}</em>2$ (g) to produce $4 \text{ moles of PCl}_3$ (g).

2. Calculating Theoretical Yield (from each reactant)

To find the theoretical yield, convert the given mass of each reactant into the mass of the product. The smaller product mass is the theoretical yield.

A. Calculation using Phosphorus ($\text{P}_4$):

• Start with given mass of P4: $45.69 \text{ g P}_4$
• Step 1: Grams of P4 to Moles of P4 (using molar mass)
  • Molar mass of P4: $4 \times (30.97 \text{ g/mol P}) = 123.88 \text{ g/mol P}_4$. (Note: Units are grams per mole, not just grams.)
  • Conversion: $\frac{1 \text{ mol P}<em>4}{123.88 \text{ g P}</em>4}$
• Step 2: Moles of P4 to Moles of PCl3 (using balanced equation mole ratio)
  • From balanced equation: $1 \text{ mol P}<em>4$ produces $4 \text{ mol PCl}</em>3$.
  • Conversion: $\frac{4 \text{ mol PCl}<em>3}{1 \text{ mol P}</em>4}$
• Step 3: Moles of PCl3 to Grams of PCl3 (using molar mass)
  • Molar mass of PCl3: $(30.97 \text{ g/mol P}) + 3 \times (35.45 \text{ g/mol Cl}) = 137.32 \text{ g/mol PCl}_3$.
  • Conversion: $\frac{137.32 \text{ g PCl}<em>3}{1 \text{ mol PCl}</em>3}$
• Calculation: $45.69 \text{ g P}<em>4 \times \frac{1 \text{ mol P}</em>4}{123.88 \text{ g P}<em>4} \times \frac{4 \text{ mol PCl}</em>3}{1 \text{ mol P}<em>4} \times \frac{137.32 \text{ g PCl}</em>3}{1 \text{ mol PCl}<em>3} = 202.6 \text{ g PCl}</em>3$
  • Result: $202.6 \text{ g of PCl}_3$ (with 4 significant figures like the initial P4 mass).

B. Calculation using Chlorine ($\text{Cl}_2$):

• Start with given mass of Cl2: $131.3 \text{ g Cl}_2$
• Step 1: Grams of Cl2 to Moles of Cl2
  • Molar mass of Cl2: $2 \times (35.45 \text{ g/mol Cl}) = 70.90 \text{ g/mol Cl}_2$.
  • Conversion: $\frac{1 \text{ mol Cl}<em>2}{70.90 \text{ g Cl}</em>2}$
• Step 2: Moles of Cl2 to Moles of PCl3
  • From balanced equation: $6 \text{ mol Cl}<em>2$ produces $4 \text{ mol PCl}</em>3$.
  • Conversion: $\frac{4 \text{ mol PCl}<em>3}{6 \text{ mol Cl}</em>2}$
• Step 3: Moles of PCl3 to Grams of PCl3 (same as above)
  • Molar mass of PCl3: $137.32 \text{ g/mol PCl}_3$.
  • Conversion: $\frac{137.32 \text{ g PCl}<em>3}{1 \text{ mol PCl}</em>3}$
• Calculation: $131.3 \text{ g Cl}<em>2 \times \frac{1 \text{ mol Cl}</em>2}{70.90 \text{ g Cl}<em>2} \times \frac{4 \text{ mol PCl}</em>3}{6 \text{ mol Cl}<em>2} \times \frac{137.32 \text{ g PCl}</em>3}{1 \text{ mol PCl}<em>3} = 169.5 \text{ g PCl}</em>3$
  • Result: $169.5 \text{ g of PCl}_3$ (with 4 significant figures).

Answers to Questions 1 and 2

• 1. Theoretical Yield: The smaller of the two calculated product masses. So, the theoretical yield is $169.5 \text{ g PCl}_3$.
• 2. Limiting Reactant: The reactant that produces the theoretical yield. In this case, chlorine gas ($\text{Cl}_2$) is the limiting reactant.
  • Note: The other reactant ($\text{P}_4$) is always the excess reactant.

3. Calculating Percent Yield

• Formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$
• Actual Yield: Given in the problem as $76.3 \text{ g}$.
• Theoretical Yield: Calculated as $169.5 \text{ g}$.
• Calculation: $\frac{76.3 \text{ g}}{169.5 \text{ g}} \times 100\% = 45.0\%$
  • Result: $45.0\%$ (with 3 significant figures, matching the actual yield).

4. Calculating Excess Reagent Remaining

• Concept: The limiting reactant is fully consumed. We use its initial amount to determine how much of the excess reactant was consumed. The difference between the initial excess reactant and the consumed excess reactant is what remains.
• Start with initial mass of Limiting Reactant (Cl2): $131.3 \text{ g Cl}_2$
• Step 1: Grams of Cl2 to Moles of Cl2
  • Molar mass of Cl2: $70.90 \text{ g/mol Cl}_2$.
  • Conversion: $\frac{1 \text{ mol Cl}<em>2}{70.90 \text{ g Cl}</em>2}$
• Step 2: Moles of Cl2 to Moles of Excess Reactant (P4) (using balanced equation mole ratio)
  • From balanced equation: $6 \text{ mol Cl}<em>2$ reacts with $1 \text{ mol P}</em>4$.
  • Conversion: $\frac{1 \text{ mol P}<em>4}{6 \text{ mol Cl}</em>2}$
• Step 3: Moles of P4 to Grams of P4 (mass of P4 consumed)
  • Molar mass of P4: $123.88 \text{ g/mol P}_4$.
  • Conversion: $\frac{123.88 \text{ g P}<em>4}{1 \text{ mol P}</em>4}$
• Calculation of P4 consumed: $131.3 \text{ g Cl}<em>2 \times \frac{1 \text{ mol Cl}</em>2}{70.90 \text{ g Cl}<em>2} \times \frac{1 \text{ mol P}</em>4}{6 \text{ mol Cl}<em>2} \times \frac{123.88 \text{ g P}</em>4}{1 \text{ mol P}<em>4} = 38.24 \text{ g P}</em>4$
  • Result: $38.24 \text{ g P}_4$ was consumed.
• Calculate P4 Remaining:
  • Initial amount of P4: $45.69 \text{ g P}_4$
  • Portion consumed: $38.24 \text{ g P}_4$
  • Remaining: $45.69 \text{ g} - 38.24 \text{ g} = 7.45 \text{ g P}_4$
  • Result: $7.45 \text{ g P}_4$ remains.

Second Stoichiometry Problem (from Textbook Example)

Problem Setup

• $2.0 \text{ grams}$ of aluminum (Al) reacts with $2.0 \text{ grams}$ of chlorine gas ($\text{Cl}<em>2$) to make aluminum chloride ($\text{AlCl}</em>3$).
• $1.75 \text{ grams}$ of $\text{AlCl}_3$ is produced (actual yield).
• Questions: Theoretical yield, limiting reactant, percent yield, excess reagent remaining.

1. Writing and Balancing the Chemical Equation

• Reactants: Aluminum (Al), Chlorine gas ($\text{Cl}_2$).
• Product: Aluminum chloride ($\text{AlCl}_3$).
• Unbalanced Equation: $\text{Al}(s) + \text{Cl}<em>2(g) \rightarrow \text{AlCl}</em>3(s)$
• Balancing Steps:
  • Balance Cl: Two Cl on left, three on right. Least common multiple is six. Add 3 to $\text{Cl}<em>2$ and 2 to $\text{AlCl}</em>3$.
    $\text{Al}(s) + 3\text{Cl}<em>2(g) \rightarrow 2\text{AlCl}</em>3(s)$
  • Balance Al: Now two Al on right. Add 2 to Al on left.
    $2\text{Al}(s) + 3\text{Cl}<em>2(g) \rightarrow 2\text{AlCl}</em>3(s)$
• Balanced Chemical Equation:
  • $2 \text{ moles of Al}$ (s) reacts with $3 \text{ moles of Cl}<em>2$ (g) to produce $2 \text{ moles of AlCl}</em>3$ (s).
  • Molar mass of Al: $26.98 \text{ g/mol}$.
  • Molar mass of Cl2: $70.90 \text{ g/mol}$.
  • Molar mass of AlCl3: $(26.98 \text{ g/mol Al}) + 3 \times (35.45 \text{ g/mol Cl}) = 133.33 \text{ g/mol AlCl}_3$.

2. Calculating Theoretical Yield

A. Calculation using Aluminum (Al):

$2.0 \text{ g Al} \times \frac{1 \text{ mol Al}}{26.98 \text{ g Al}} \times \frac{2 \text{ mol AlCl}_3}{2 \text{ mol Al}} \times \frac{133.33 \text{ g AlCl}_3}{1 \text{ mol AlCl}_3} = 9.88 \text{ g AlCl}_3$
*   Result: $9.88 \text{ g AlCl}_3$ (with 3 significant figures).

B. Calculation using Chlorine ($\text{Cl}_2$):

$2.0 \text{ g Cl}_2 \times \frac{1 \text{ mol Cl}_2}{70.90 \text{ g Cl}_2} \times \frac{2 \text{ mol AlCl}_3}{3 \text{ mol Cl}_2} \times \frac{133.33 \text{ g AlCl}_3}{1 \text{ mol AlCl}_3} = 2.51 \text{ g AlCl}_3$
*   Result: $2.51 \text{ g AlCl}_3$ (with 3 significant figures).

Answers to Questions 1 and 2

• 1. Theoretical Yield: $2.51 \text{ g AlCl}_3$.
• 2. Limiting Reactant: Chlorine gas ($\text{Cl}_2$).
  • Note: Aluminum (Al) is the excess reactant.

3. Calculating Percent Yield

• Actual Yield: $1.75 \text{ g}$.
• Theoretical Yield: $2.51 \text{ g}$.
• Calculation: $\frac{1.75 \text{ g}}{2.51 \text{ g}} \times 100\% = 69.7\%$
  • Result: $69.7\%$ (with 3 significant figures).

4. Calculating Excess Reagent Remaining (Aluminum)

• Start with initial mass of Limiting Reactant (Cl2): $2.0 \text{ g Cl}_2$
• Step 1: Grams of Cl2 to Moles of Cl2
  • Conversion: $\frac{1 \text{ mol Cl}<em>2}{70.90 \text{ g Cl}</em>2}$
• Step 2: Moles of Cl2 to Moles of Excess Reactant (Al)
  • Conversion: $\frac{2 \text{ mol Al}}{3 \text{ mol Cl}_2}$
• Step 3: Moles of Al to Grams of Al (mass of Al consumed)
  • Conversion: $\frac{26.98 \text{ g Al}}{1 \text{ mol Al}}$
• Calculation of Al consumed: $2.0 \text{ g Cl}<em>2 \times \frac{1 \text{ mol Cl}</em>2}{70.90 \text{ g Cl}<em>2} \times \frac{2 \text{ mol Al}}{3 \text{ mol Cl}</em>2} \times \frac{26.98 \text{ g Al}}{1 \text{ mol Al}} = 0.507 \text{ g Al}$
  • Result: $0.507 \text{ g Al}$ was consumed.
• Calculate Al Remaining:
  • Initial amount of Al: $2.0 \text{ g Al}$.
  • Portion consumed: $0.507 \text{ g Al}$.
  • Remaining: $2.0 \text{ g} - 0.507 \text{ g} = 1.49 \text{ g Al}$.
  • Result: $1.49 \text{ g Al}$ remains.

Chapter 5: Introduction to Solutions and Aqueous Reactions

Solutions: Homogeneous Mixtures

• Definition: A solution is a homogeneous mixture where components are uniformly distributed.
  • Example: Salt dissolving in water. The salt is still present, but visually, it's uniform.
• Homogeneous Mixtures: Have the same ratio of components regardless of the sample taken.
  • Shorthand: