Stoichiometry: Limiting Reactants, Yields, and Thermochemical Energy

Fundamental Concepts of Limiting and Excess Reactants

• Excess Reactant: This is the reactant that remains after the chemical reaction has reached completion. It is defined as the substance that is "more than is needed" and is left over because there is no more of the alternate reactant to continue the reaction.
• Limiting Reactant: This is the reactant present in the smallest stoichiometric amount, which limits the formation of the product. It is totally consumed by the end of the reaction, leaving nothing behind.
• Product Formation: The amount of product generated in a reaction is strictly determined by the limiting reactant, never by the excess reactant. The excess reactant simply sits unconsumed alongside the product at the end of the process.

The Pancake Analogy for Stoichiometry

• Stoichiometry is often compared to a recipe in baking or cooking, where ingredients are reactants and the final dish is the product.
• Original Recipe (Standard Ratio):
  • $1\text{ cup of flour}$
  • $2\text{ eggs}$
  • Baking powder (amount implied as part of the ratio)
  • Product: $5\text{ pancakes}$
• Scenario for Comparison:
  • Available Ingredients: $3\text{ cups of flour}$, $10\text{ eggs}$, and "four times the amount" of baking powder.
  • Potential Yield from Flour: $\frac{3\text{ cups flour}}{1\text{ cup flour}} \times 5\text{ pancakes} = 15\text{ pancakes}$
  • Potential Yield from Eggs: $\frac{10\text{ eggs}}{2\text{ eggs}} \times 5\text{ pancakes} = 25\text{ pancakes}$
  • Potential Yield from Baking Powder: Assuming the original was $1$ unit, $4 \times 5\text{ pancakes} = 20\text{ pancakes}$ (though the speaker mentions a potential for $40$ in a different context, the logic remains finding the lowest).
• Determining the Outcome: The real amount of product is determined by the lowest calculated number ($15$ pancakes). This is because once the flour is consumed to make $15$ pancakes, there is no flour left to react with the remaining eggs and baking powder, regardless of their quantity.

Yield Definitions and Calculations

• Theoretical Yield: The maximum amount of product that can be generated, as determined by calculation from the limiting reactant. In the pancake scenario, the theoretical yield is $15$ pancakes.
• Actual Yield: The amount of product actually produced during an experiment or process. This is often less than the theoretical yield due to spills, incomplete reactions, or experimental errors (e.g., dropping $4$ pancakes on the floor, resulting in an actual yield of $11$).
• Percent Yield Formula:
  • $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$
• Calculation Example: If actual yield is $11$ and theoretical is $15$:
  • $\frac{11}{15} \times 100 = 73.3\%$

Determining Limiting Reactant and Theoretical Yield: Method 1

This method involves calculating the amount of product obtained from each given amount of reactant and choosing the smallest result.

• Example Problem: Titanium ($Ti$) and Chlorine ($Cl_2$) reacting to form Titanium(IV) chloride ($TiCl_4$).
• Given Data:
  • $1.8\text{ moles of titanium}$
  • $3.2\text{ moles of chlorine}$
• Calculation from Titanium:
  • Based on the balanced ratio of $1:1$:
  • $1.8\text{ moles } Ti \times \frac{1\text{ mole } TiCl_4}{1\text{ mole } Ti} = 1.8\text{ moles } TiCl_4$
• Calculation from Chlorine:
  • Based on the balanced ratio of $1\text{ mole } TiCl_4$ to $2\text{ moles } Cl_2$:
  • $3.2\text{ moles } Cl_2 \times \frac{1\text{ mole } TiCl_4}{2\text{ moles } Cl_2} = 1.6\text{ moles } TiCl_4$
• Outcome:
  • Theoretical Yield: $1.6\text{ moles } TiCl_4$ (the smallest result).
  • Limiting Reactant: Chlorine ($Cl_2$), because it produces the smaller amount of product.

Identifying Limiting and Excess Reactants: Method 2

This method compares the reactants to each other using their stoichiometric ratio to see which one runs out first, without necessarily calculating the product yield.

• Example Problem: Calcium nitrate ($Ca(NO_3)_2$) reacting with Lithium phosphate ($Li_3PO_4$).
• Balanced Equation Ratio: $3\text{ moles } Ca(NO_3)_2 : 2\text{ moles } Li_3PO_4$
• Given Data:
  • $3.4\text{ moles of calcium nitrate}$
  • $2.4\text{ moles of lithium phosphate}$
• Calculation Check A: To react with $3.4\text{ moles of calcium nitrate}$, how much lithium phosphate is needed?
  • $3.4\text{ moles } Ca(NO_3)_2 \times \frac{2\text{ moles } Li_3PO_4}{3\text{ moles } Ca(NO_3)_2} = 2.266\text{ moles } Li_3PO_4$
  • Comparison: We have $2.4$ moles available and only need $2.266$. Therefore, lithium phosphate is in excess.
• Calculation Check B: To react with $2.4\text{ moles of lithium phosphate}$, how much calcium nitrate is needed?
  • $2.4\text{ moles } Li_3PO_4 \times \frac{3\text{ moles } Ca(NO_3)_2}{2\text{ moles } Li_3PO_4} = 3.6\text{ moles } Ca(NO_3)_2$
  • Comparison: We have $3.4$ moles available but need $3.6$. Therefore, calcium nitrate is the limiting reactant.

Multi-Step Stoichiometry with Grams and Percent Yield

Real-world problems often provide mass in grams, requiring conversion to moles before using stoichiometric ratios.

• Reaction: $2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$
• Given Data:
  • Mass of Sodium ($Na$): $53.2\,g$
  • Mass of Chlorine ($Cl_2$): $65.8\,g$
  • Actual Yield of $NaCl$: $86.4\,g$
• Step 1: Convert Reactants to Moles:
  • Sodium: $\frac{53.2\,g}{23.0\,g/mol} = 2.313\text{ moles } Na$
  • Chlorine: $\frac{65.8\,g}{70.9\,g/mol} = 0.928\text{ moles } Cl_2$
• Step 2: Calculate Theoretical Mass of Product (NaCl):
  • From $53.2\,g\,Na$: $2.313\text{ moles } Na \times \frac{2\text{ moles } NaCl}{2\text{ moles } Na} \times 58.44\,g/mol = 135.17\,g\,NaCl$
  • From $65.8\,g\,Cl_2$: $0.928\text{ moles } Cl_2 \times \frac{2\text{ moles } NaCl}{1\text{ mole } Cl_2} \times 58.44\,g/mol = 108.47\,g\,NaCl$
• Step 3: Determine Theoretical Yield and Limiting Reactant:
  • Theoretical Yield: $108.47\,g$ (the smaller value).
  • Limiting Reactant: Chlorine ($Cl_2$).
• Step 4: Calculate Percent Yield:
  • $\text{Percent Yield} = \frac{86.4\,g}{108.47\,g} \times 100 \approx 79.6\%$

Energy and Enthalpy in Chemical Reactions

Chemical reactions are accompanied by energy changes, which can be quantified through stoichiometry.

• Enthalpy of Reaction ($\Delta H_{rxn}$): This represents the total amount of energy absorbed or produced by a reaction at constant pressure. It is a state function, meaning it depends on the final state (products) and initial state (reactants).
• Exothermic Processes:
  • Energy is produced/released.
  • $\Delta H$ is negative.
  • Energy can be treated as a product in the reaction.
  • Example: Methane combustion ($CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$) where $\Delta H = -802.3\,kJ$.
• Endothermic Processes:
  • Energy is absorbed/added to reactants.
  • $\Delta H$ is positive.
  • Energy can be treated as a reactant.
  • Example: Nitrogen and Oxygen reaction where $\Delta H = +182.6\,kJ$.
• Stoichiometry with Energy:
  • Ratios can be established between the stoichiometric coefficients (moles) and the enthalpy ($\Delta H$).
  • For methane: $\frac{1\text{ mole } CH_4}{-802.3\,kJ}$ or $\frac{-802.3\,kJ}{1\text{ mole } CH_4}$.
  • If the amount of reactant is doubled, the energy produced also doubles, but the ratio remains constant.

Enthalpy Calculation Example: Propane Combustion

• Reaction: Propane combustion ($C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$).
• Given Energy: $\Delta H = -2044\,kJ$ per $1\text{ mole } C_3H_8$.
• Mass of Propane: $11.8 \times 10^{3}\,g$ (which is $11.8\,kg$).
• Objective: Find the total energy produced in kilojoules.
• Steps:
  1. Convert mass to moles using molar mass of propane ($C_3H_8$).
  2. Use the energy-to-mole ratio: $\frac{-2044\,kJ}{1\text{ mole } C_3H_8}$.
  3. Multiply the number of moles by the energy ratio to find total heat produced.

Questions & Discussion

• Student Question: "Because it's not just asking for one, does it not matter what one we put on top this time of the ratio?"
• Instructor Response: "It matters. I'm gonna show you because this time, we have to work only with this mole ratio between reactants. Nothing about the products… how I'm gonna write this mole ratio with what I need on the top… you have to divide by the molar mass."
• Student Observation on Percent Yield: "Can we get more than 100 as percent yield?"
• Instructor Response: "Sometimes it happens, but it's for sure… we don't have totally dried product, and we have, like, water inside. Something was wrong with the procedure. Percent yield greater than 100 means something is wrong."