Limiting Reagents, Yield, and Stoichiometry Notes

Definition of Acids: - Substances that provide $H^+$ ions (hydrogen ions) when dissolved in water are classified as acids. - The $H^+$ cation is described as a 'bare proton' because it possesses no electrons ( $e^-$ ).
Acids in Aqueous Solution: - In aqueous environments, the $H^+$ ion do not exist in isolation; it associates with water ( $H_2O$ ) to form the hydronium ion, denoted as $H_3O^+(aq)$ . - The terms $H_3O^+(aq)$ and $H^+(aq)$ are often used interchangeably in chemical equations.
Polarization in Acidic Molecules: - An example provided is Hydrogen Chloride ( $HCl(g)$ ), which contains a highly polarized covalent bond between the Hydrogen and Chlorine atoms. - When placed in water, this bond ionizes to produce $H^+(aq)$ and $Cl^-(aq)$ .

Reaction with Metals: - General Word Equation: $\text{Metal} + \text{Acid} \rightarrow \text{Salt} + \text{Hydrogen}$ . - Formula Equation: $Zn + 2HCl \rightarrow ZnCl_2 + H_2$ . - Complete Ionic Equation: $Zn(s) + 2H^+(aq) + 2Cl^-(aq) \rightarrow Zn^{2+}(aq) + H_2(g) + 2Cl^-(aq)$ . - Net Ionic Equation: $Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g)$ . - Isolation of salt: The solid salt can be isolated by evaporating the solvent: $Zn^{2+}(aq) + 2Cl^-(aq) \rightarrow ZnCl_2(s)$ .
Reaction with Carbonates: - General Word Equation: $\text{Carbonate} + \text{Acid} \rightarrow \text{Salt} + H_2O + CO_2$ . - Formula Equation: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$ . - Complete Ionic Equation: $CaCO_3(s) + 2H^+(aq) + 2Cl^-(aq) \rightarrow Ca^{2+}(aq) + H_2O(l) + CO_2(g) + 2Cl^-(aq)$ . - Net Ionic Equation: $CaCO_3(s) + 2H^+(aq) \rightarrow Ca^{2+}(aq) + H_2O(l) + CO_2(g)$ . - Isolation of salt: The solid calcium chloride can be isolated via evaporation: $Ca^{2+}(aq) + 2Cl^-(aq) \rightarrow CaCl_2(s)$ .

Relative Atomic Mass: This is the weighted average mass of all naturally occurring isotopes of a specific element relative to the isotope ${}^{12}C$ .
Avogadro Number: Represented as $6.022 \times 10^{23}$ . This specific value is defined as the number of atoms of ${}^{12}C$ found in exactly $12.00\,g$ of ${}^{12}C$ .
The Mole: The term used to represent an Avogadro number of objects (atoms, molecules, ions, etc.).
The Molar Mass Formula: - $\text{Amount in moles (n)} = \frac{\text{mass (m)}}{\text{molar mass (M)}}$ - Expressed as: $n = \frac{m}{M}$ .

Problem Scenario: Determining the masses of Sodium ( $Na$ ) and Chlorine ( $Cl_2$ ) required to produce $100\,g$ of Sodium Chloride ( $NaCl$ ).
Balanced Equation: $2Na + Cl_2 \rightarrow 2NaCl$ (or simplified as $Na + \frac{1}{2}Cl_2 \rightarrow NaCl$ ).
Calculations for $100\,g$ of $NaCl$ : - Molar Mass of $NaCl$ ( $M$ ): $58.5\,g\,mol^{-1}$ . - Moles of $NaCl$ ( $n$ ): $\frac{100\,g}{58.5\,g\,mol^{-1}} = 1.71\,mol$ . - Molar ratio requirement: For every $1\,mol$ of $NaCl$ , $1\,mol$ of $Na$ and $0.5\,mol$ of $Cl_2$ are needed. - Moles of $Na$ needed: $1.71\,mol$ . - Moles of $Cl_2$ needed: $0.855\,mol$ . - Calculation of Mass ( $m = n \times M$ ): - Mass of $Na$ : $1.71\,mol \times 23.0\,g\,mol^{-1} = 39.3\,g$ . - Mass of $Cl_2$ : $0.855\,mol \times 70.9\,g\,mol^{-1} = 60.6\,g$ .

Conceptual Definition: In most chemical reactions, reagents are not present in exact stoichiometric proportions. The reagent that is entirely consumed first is the limiting reagent.
Function: The limiting reagent determines the maximum amount of product that can be formed ("the theoretical yield").
Example Analysis: Reaction of Hydrogen ( $H_2$ ) and Fluorine ( $F_2$ ): - Equation: $H_2 + F_2 \rightarrow 2HF$ . - Initial Conditions: $5.00\,g$ of $H_2$ and $10.0\,g$ of $F_2$ . - Steps to determine limiting reagent: - Moles of $H_2$ : $\frac{5.00\,g}{2.01\,g\,mol^{-1}} = 2.49\,mol$ . - Moles of $F_2$ : $\frac{10.0\,g}{38.0\,g\,mol^{-1}} = 0.263\,mol$ . - Comparison: Since the reaction requires a $1:1$ ratio, $F_2$ is the limiting reagent because there is much less of it available ( $0.263\,mol$ ) compared to $H_2$ ( $2.49\,mol$ ). - Final Mass Balance: - Moles of $F_2$ used: $0.263\,mol$ (remains: $0\,g$ ). - Moles of $H_2$ used: $0.263\,mol$ (remains: $2.49 - 0.263 = 2.23\,mol$ ). - Mass of $H_2$ remaining: $2.23\,mol \times 2.01\,g\,mol^{-1} = 4.48\,g$ . - Moles of $HF$ produced: $2 \times 0.263 = 0.526\,mol$ . - Mass of $HF$ produced: $0.526\,mol \times 20.0\,g\,mol^{-1} = 10.5\,g$ .

Condition of Completion: 100% reaction occurs when all of the limiting reagent is used up. However, reactions often do not go to completion.
Formula for % Yield: - $\text{\% yield} = \frac{\text{actual mass of product}}{\text{max theoretical mass}} \times 100$
Example Calculation: - Reaction: $5.00\,g$ of Sodium ( $Na$ ) reacts with excess Chlorine ( $Cl_2$ ) to produce $12.4\,g$ of Sodium Chloride ( $NaCl$ ). - Theoretical Yield Calculation: - Moles of $Na$ : $\frac{5.00\,g}{23.0\,g\,mol^{-1}} = 0.217\,mol$ . - Stoichiometric moles of $NaCl$ expected (from 1:1 ratio with $Na$ ): $0.217\,mol$ . - Theoretical mass of $NaCl$ : $0.217\,mol \times 58.5\,g\,mol^{-1} = 12.7\,g$ . - Yield Calculation: - $\text{\% yield} = \frac{12.4\,g}{12.7\,g} \times 100 = 97.6\%$

Structural Chemistry: Distinguish between molecular and network covalent solids.
Bonding: Predict polar covalent bonds based on differences in nuclear charges.
Formulae: Use molecular and empirical formulae correctly in various contexts.
Equations: Identify products/reactants and balance both atoms and charges.
Acids: Describe the ionization of acids in water to produce $H^+$ ions.
Stoichiometry Skills: - Determine mass of product from mass of starting materials (and vice versa). - Identify the limiting reagent and apply it to calculations. - Determine percentage yield based on experimental vs. theoretical masses.

Problem 1: Write a balanced equation for the complete combustion of carbon to form carbon dioxide ( $C + O_2 \rightarrow CO_2$ ).
Problem 2: Calculate the number of moles of carbon in $100\,g$ of carbon.
Problem 3: Identify the limiting reagent if the combustion is carried out in open air (where oxygen is in excess).
Problem 4: Calculate the moles of $CO_2$ produced from complete combustion of $100\,g$ of carbon.
Problem 5: Determine the mass of $CO_2$ produced from the complete combustion of $100\,g$ of carbon.
Problem 6: Calculate the percentage yield if only $348\,g$ of $CO_2$ is isolated from the specified $100\,g$ carbon reaction.