Comprehensive Notes on Stoichiometry, Limiting Reagents, and Enthalpy

Advanced Stoichiometry and Molar Conversions

• The standard process for stoichiometry involves moving from mass to moles, moles to moles, and then moles back to mass:
  • $\\text{Grams of Species A} \\rightarrow \\text{Moles of Species A} \\rightarrow \\text{Moles of Species B} \\rightarrow \\text{Grams of Species B}$
• Common Pitfall: A frequent error in chemical calculations is failing to consider stoichiometric coefficients during the conversion from grams to moles.
• Conversion factor roles:
  • Use Molar Mass to convert between mass (grams) and moles.
  • Use Stoichiometric Coefficients exclusively when converting from moles of one species to moles of a different species.

The Limiting Reagent Concept

• When specific amounts of multiple reagents are provided, one will likely be used up before the others. This is the Limiting Reagent.
• Bicycle Metaphor:
  • To build one bicycle, you need 1 frame and 2 wheels.
  • If you have 4 frames and 4 wheels, you can only make 2 bicycles.
  • Reason: You run out of wheels. The wheels are the limiting reagent. The frames are in excess (2 frames will be left over).
• Definition: The limiting reagent is the chemical species that runs out first in a reaction, thereby limiting the maximum amount of product that can be formed.

Identifying the Limiting Reagent

• You cannot determine the limiting reagent simply by looking at the starting masses or mole amounts.
• Method: Perform stoichiometry for every reagent to calculate how much of a single product each could theoretically produce.
• Example: 12 moles of $H_2$ and 7 moles of $O_2$ reacting to form $H_2O$.
  • Stoichiometry for $H_2$: $12\\,\\text{moles H}_2 \\times \\frac{2\\,\\text{moles H}_2O}{2\\,\\text{moles H}_2} = 12\\,\\text{moles H}_2O$
  • Stoichiometry for $O_2$: $7\\,\\text{moles O}_2 \\times \\frac{2\\,\\text{moles H}_2O}{1\\,\\text{mole O}_2} = 14\\,\\text{moles H}_2O$
• Result: $H_2$ is the limiting reagent because it produces the least amount of product (12 moles versus 14 moles).
• The maximum amount of product possible is the lower value (12 moles), while the excess reagent ($O_2$) remains unreacted.

Calculation Example: Synthesis of Lithium Fluoride

Problem: Determine the maximum amount of lithium fluoride produced from $10.0\\,\\text{g}$ of lithium and $10.0\\,\\text{g}$ of fluorine gas.

• Step 1: Balance the equation.
  • $2Li(s) + F_2(g) \\rightarrow 2LiF(s)$
• Step 2: Calculate product from Lithium (Li).
  • $10.0\\,\\text{g Li} \\times \\frac{1\\,\\text{mole Li}}{6.94\\,\\text{g Li}} \\times \\frac{2\\,\\text{moles LiF}}{2\\,\\text{moles Li}} \\times \\frac{25.94\\,\\text{g LiF}}{1\\,\\text{mole LiF}} = 37.4\\,\\text{g LiF}$
• Step 3: Calculate product from Fluorine ($F_2$).
  • Note: Fluorine is a diatomic molecule ($F_2$). Its molar mass is $2 \\times 18.99 = 37.98\\,\\text{g/mol}$.
  • $10.0\\,\\text{g F}_2 \\times \\frac{1\\,\\text{mole F}_2}{37.98\\,\\text{g F}_2} \\times \\frac{2\\,\\text{moles LiF}}{1\\,\\text{mole F}_2} \\times \\frac{25.94\\,\\text{g LiF}}{1\\,\\text{mole LiF}} = 13.7\\,\\text{g LiF}$
• Conclusion: Fluorine is the limiting reagent. The maximum yield is $13.7\\,\\text{g LiF}$.

Determining Excess Reagent Remaining

• To find how much excess reagent is left over, you must determine how much was consumed.
• Calculation Process:
  1. Perform stoichiometry from the limiting reagent to the excess reagent amount.
  2. Subtract the amount consumed from the original amount given.
• Using the LiF example:
  • Initial Lithium: $10.0\\,\\text{g}$.
  • Stoichiometry from limiting $F_2$ ($10.0\\,\\text{g}$) to lithium consumed:
  • $10.0\\,\\text{g F}_2 \\times \\frac{1\\,\\text{mole F}_2}{37.98\\,\\text{g F}_2} \\times \\frac{2\\,\\text{moles Li}}{1\\,\\text{mole F}_2} \\times \\frac{6.94\\,\\text{g Li}}{1\\,\\text{mole Li}} = 3.65\\,\\text{g Li consumed}$
  • Lithium remaining: $10.0\\,\\text{g} - 3.65\\,\\text{g} = 6.35\\,\\text{g}$.
• Significant Figures: In subtraction, the result is determined by the least number of decimal places. ($10.0$ has one decimal place, $3.65$ has two, so the answer must have one: $6.4\\,\\text{g}$, though the transcript notes $6.35\\,\\text{g}$ initially).

Percent Yield

• Percent yield compares the actual result obtained in a lab to the theoretical result calculated via stoichiometry.
• Formula: $\\text{Percent Yield} = \\frac{\\text{Actual Yield}}{\\text{Theoretical Yield}} \\times 100$
• Applied Example: If lab results yield $4.53\\,\\text{g LiF}$ and the theoretical yield was $13.7\\,\\text{g}$, the calculation is:
  • $\\frac{4.53\\,\\text{g}}{13.7\\,\\text{g}} \\times 100 = 33.1\\%$ yield.

Chemical Enthalpy (Heat in Reactions)

• Enthalpy (\Delta H): The amount of thermal energy released or absorbed by a chemical reaction.
• Exothermic Reactions:
  • Heat is released from the system.
  • Heat can be treated as a product in the equation.
  • $\\Delta H$ is negative ($-\\Delta H$).
• Endothermic Reactions:
  • Heat is absorbed into the system.
  • Heat can be treated as a reagent.
  • $\\Delta H$ is positive ($+\\Delta H$).
• Flipping Reactions: If a reaction is reversed, the magnitude of enthalpy remains the same, but the sign changes.
  • Example: Combustion of propane ($\\Delta H = -2044\\,\\text{kJ}$). Reversing it to form propane from $CO_2$ and $H_2O$ results in $\\Delta H = +2044\\,\\text{kJ}$.

Stoichiometry of Enthalpy

• Enthalpy values are proportional to the stoichiometric coefficients in a balanced equation and can be used as conversion factors.
• Example: Combustion of propane ($C_3H_8$):
  • Molar mass of propane: $44.1\\,\\text{g/mol}$.
  • Given: $1.18 \\times 10^4\\,\\text{g}$ of propane.
  • Calculation: $1.18 \\times 10^4\\,\\text{g C}_3H_8 \\times \\frac{1\\,\\text{mole C}_3H_8}{44.1\\,\\text{g C}_3H_8} \\times \\frac{-2044\\,\\text{kJ}}{1\\,\\text{mole C}_3H_8} = -5.47 \\times 10^5\\,\\text{kJ}$.
  • Interpretation: The negative sign or the phrase "was released" both indicate the loss of heat from the system.

Balanced Reaction and States: Diboron Trioxide

• Reaction: Boron powder ($B(s)$) + Oxygen gas ($O_2(g)$) $\\rightarrow$ Diboron trioxide ($B_2O_3(s)$).
• Balancing Technique: "Even-Odd Split". Since Oxygen is $O_2$ on the left and $O_3$ on the right, multiply species to normalize.
  • Balanced Equation: $4B(s) + 3O_2(g) \\rightarrow 2B_2O_3(s)$.
  • Given enthalpy ($\\Delta H$) for this reaction is $-1237\\,\\text{kJ}$.
• Calculation for $185\\,\\text{g B}_2O_3$:
  • Molar mass of $B_2O_3$: $69.62\\,\\text{g/mol}$.
  • Calculation: $185\\,\\text{g B}_2O_3 \\times \\frac{1\\,\\text{mole B}_2O_3}{69.62\\,\\text{g B}_2O_3} \\times \\frac{-1237\\,\\text{kJ}}{2\\,\\text{moles B}_2O_3} = -1640\\,\\text{kJ}$ (rounded to 3 sig figs: $-1640\\,\\text{kJ}$ or $1640\\,\\text{kJ}$ released).

Stoichiometry with Gas Laws

• Stoichiometry can be combined with Ideals Gas Laws ($PV = nRT$) to find gas volumes.
• Stoichiometry at STP (Standard Temperature and Pressure):
  • STP Pressure: $1\\,\\text{atm}$.
  • STP Temperature: $273\\,\\text{K}$ (or $273.15\\,\\text{K}$ for more precision).
• Example: Producing $CO_2$ from $500\\,\\text{g}$ Carbon:
  • Reaction: $C(s) + O_2(g) \\rightarrow CO_2(g)$.
  • $500\\,\\text{g C} \\times \\frac{1\\,\\text{mole C}}{12.01\\,\\text{g C}} \\times \\frac{1\\,\\text{mole CO}_2}{1\\,\\text{mole C}} = 41.6\\,\\text{moles CO}_2$.
  • Using $V = \\frac{nRT}{P}$:
  • $V = \\frac{41.6\\,\\text{moles} \\times 0.08206 \\, \\frac{\\text{dm}^3 \\cdot \\text{atm}}{\\text{mol} \\cdot \\text{K}} \\times 273\\,\\text{K}}{1\\,\\text{atm}} = 932\\,\\text{dm}^3\\,\\text{CO}_2$.

Liter-to-Liter Conversions for Gases

• Under conditions where temperature and pressure remain constant, the stoichiometric coefficients for gaseous species can be used for direct volume-to-volume conversions.
• Avogadro's Principle: All gases behave similarly under matching conditions; therefore, the mole-to-mole ratio is equivalent to the volume-to-volume ratio.
• Example: If $9.0\\,\\text{dm}^3$ of $HCl$ is produced from $H_2$ gas:
  • Reaction: $H_2(g) + Cl_2(g) \\rightarrow 2HCl(g)$.
  • Ratio: $1\\,\\text{dm}^3\\,H_2 : 2\\,\\text{dm}^3\\,HCl$.
  • Calculation: $9.0\\,\\text{dm}^3\\,HCl \\times \\frac{1\\,\\text{dm}^3\\,H_2}{2\\,\\text{dm}^3\\,HCl} = 4.5\\,\\text{dm}^3\\,H_2$.

Questions & Discussion

• Q: Which elements are diatomic?
  • A: "I Have No Fear Of Ice Cold Beer/Fever" (mnemonic). Elements are: Hydrogen ($H_2$), Nitrogen ($N_2$), Fluorine ($F_2$), Oxygen ($O_2$), Iodine ($I_2$), Chlorine ($Cl_2$), and Bromine ($Br_2$).
• Q: Why was the house fire in East LA mentioned?
  • A: To illustrate that "combustion" in the real world is broader than simple carbon-hydrogen burning. Toxic chemicals like lead, asbestos, and chromium in materials produce dangerous combustion products. Advice: Wear a mask for at least four days if a large structure fire occurs in your neighborhood.
• Q: When do we use 273.15 versus 273 for STP?
  • A: Both are appropriate; use the level of precision (sig figs) demanded by your starting values. Do not let STP values limit the precision of your final answer.
• Q: Why use coefficients in the calculation but not in the molar mass?
  • A: Molar mass only applies to the species itself. Coefficients are only used during the mole-to-mole conversion step to reflect the ratio of the balanced equation.
• Q: What is unique about hydrocarbons in combustion?
  • A: They must contain Carbon and Hydrogen to burn effectively; other elements present (Nitrogen, Sulfur) create the toxicity in smoke.