Stoichiometry, Limiting Reagents, Yields & Solution Chemistry

Module 7: Stoichiometry - Introduction and Mole Concept

A. Introduction to Stoichiometry

1. Reaction Stoichiometry

Definition: Reaction stoichiometry uses balanced chemical equations to quantitatively determine:
- The amount of each reagent (reactant) needed.
- The amount of each product formed.
Example: The Haber Process: H2(g) + N2(g) \rightarrow NH_3(g)
- When balanced: 3H2(g) + N2(g) \rightarrow 2NH_3(g)
- This indicates a ratio of 3 molecules of Hydrogen to 1 molecule of Nitrogen producing 2 molecules of Ammonia.
- The stoichiometric ratio between Hydrogen and Ammonia is 3 \text{ to } 2.
Key Principle: In a balanced chemical equation, if the quantity of one substance (reactant or product) is known, all other quantities (mass, moles, molecules, atoms) can be determined.

2. The Mole Concept

Problem: Reactions cannot be carried out at the molecular scale due to extremely small masses.
- Example: For the reaction C2H5OH(l) + 3O2(g) \rightarrow 2CO2(g) + 3H2O(l) at the molecular scale, one molecule of C2H5OH would weigh 7.65 \times 10^{-23} \text{ g} and three molecules of O2 would weigh 15.9 \times 10^{-23} \text{ g}.
Solution: The mole is used as a standard unit of measure for convenience in chemical reactions.
Molar Mass (MM): The mass of one mole of a substance.
- MM(C2H5OH) = 46.08 \text{ g/mol}
- MM(O_2) = 32.00 \text{ g/mol}
Avogadro's Number (N_A): 6.02 \times 10^{23} \text{ molecules/mol}. This is the number of entities (atoms, molecules, ions) in one mole of a substance.
Conversion from Molar Mass to mass per molecule:
- \text{gram per molecule of } C2H5OH = \frac{46.08 \text{ g/mol}}{6.02 \times 10^{23} \text{ molecules/mol}} = 7.65 \times 10^{-23} \text{ g/molecule}

B. Determining the Number of Moles (n) in Various Systems

1. For Solids

Formula: \text{# moles} (n) = \frac{\text{mass (g)}}{\text{Molar Mass (g/mol)}}

2. For Solutions

Molarity (C_{\text{solute}}): Defined as the number of moles of solute per liter of solution.
- Formula: C_{\text{solute}} = \text{Molarity (mol/L)} = \frac{\text{Number of moles of solute (mol)}}{\text{Volume of solution (L)}}
Number of moles in a given volume of solution:
- Formula: \text{# moles} (n) = \text{Volume used (L)} \times C_{\text{solute}} \text{ (mol/L)}

3. For Gases

Ideal Gas Law: PV = nRT
Number of moles (n):
- Formula: n = \frac{P \text{ (kPa)} \times V \text{ (L)}}{R \text{ (kPa} \cdot \text{L/mol} \cdot \text{K)} \times T \text{ (K)}}
Variables and Units:
- V = Volume (units: L or dm^3)
- R = Ideal Gas Constant:
  - 8.314 \text{ J/(K} \cdot \text{mol)}
  - 8.314 \text{ kPa} \cdot \text{L/(K} \cdot \text{mol)}
  - 0.0821 \text{ atm} \cdot \text{L/(K} \cdot \text{mol)}
  - 0.0831 \text{ bar} \cdot \text{L/(K} \cdot \text{mol)}
- T = Temperature (units: Kelvin)
- P = Pressure (units: atm, Pa, or Torr)
STP (Standard Temperature & Pressure):
- T = 0^{\circ}\text{C} (273.15 \text{ K})
- P = 1 \text{ atm} (101.325 \text{ kPa})

4. For Liquids

Often, density is used to find mass, which is then converted to moles.
Formula: \text{mass (g)} = \text{density (g/mL)} \times \text{volume (mL)}
Combining with mole concept:
- \text{# moles} (n) = \frac{\text{density (g/mL)} \times \text{volume (mL)}}{\text{Molar Mass (g/mol)}}

C. Stoichiometric Calculations - Example 1

Problem: What mass of I2 is produced if 13.1 \text{ g} KI is reacted with excess KIO3 and HNO3? (Reaction: 5KI(aq) + KIO3(aq) + 6HNO3(aq) \rightarrow 6KNO3(aq) + 3I2(aq) + 3H2O(l))
Given: MM(KI) = 166.0 \text{ g/mol}, MM(I_2) = 253.8 \text{ g/mol}
Stoichiometry from balanced equation: 5 moles KI produce 3 moles I_2.
- Ratio: \frac{\text{# moles of } KI \text{ reacted}}{\text{# moles of } I2 \text{ produced}} = \frac{5 \text{ moles}}{3 \text{ moles}} or \frac{\text{# moles of } I2 \text{ produced}}{\text{# moles of } KI \text{ reacted}} = \frac{3 \text{ moles}}{5 \text{ moles}}
- Tip: Always use the expression that places the number of moles you are looking for in the numerator.
Steps:
1. Convert mass of KI to moles of KI:
  - n_{KI} = \frac{13.1 \text{ g}}{166.0 \text{ g/mol}} = 7.8916 \times 10^{-2} \text{ mol KI}
2. Convert moles of KI to moles of I_2 (using stoichiometry):
  - n{I2 \text{ produced}} = \frac{3 \text{ moles } I2}{5 \text{ moles } KI} \times n{KI} = \frac{3}{5} \times 7.8916 \times 10^{-2} \text{ mol} = 4.7349 \times 10^{-2} \text{ mol } I_2
3. Convert moles of I2 to mass of I2:
  - m{I2} = n{I2 \text{ produced}} \times MM(I_2) = 4.7349 \times 10^{-2} \text{ mol} \times 253.8 \text{ g/mol} = 12.0 \text{ g} (rounded to 3 significant figures)

D. Stoichiometric Calculations - Example 2

Reaction: Na2CO3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2O(l) + CO2(g)
Reactants:
- Na2CO3: m = 2.0 \text{ g}, MM = 105.99 \text{ g} \cdot \text{mol}^{-1}. Diluted in 20 \text{ mL} water.
- HCl: Aqueous solution with Molarity 1 \text{ mol} \cdot \text{L}^{-1}.

1. What volume of HCl is needed to neutralize the sodium carbonate?

*   **1.a. Number of moles of Na_2CO_3:**
    *   n_{Na_2CO_3} = \frac{2.0 \text{ g}}{105.99 \text{ g} \cdot \text{mol}^{-1}} = 0.019 \text{ mol}
*   **1.b. Number of moles of HCl needed (stoichiometry):**
    *   From the equation: n_{HCl} = 2 \times n_{Na_2CO_3}
    *   n_{HCl} = 2 \times 0.019 \text{ mol} = 0.038 \text{ mol}
*   **1.c. Volume of HCl solution needed:**
    *   Using n = C \times V: V_{HCl} = \frac{n_{HCl}}{C_{HCl}} = \frac{0.038 \text{ mol}}{1 \text{ mol} \cdot \text{L}^{-1}} = 0.038 \text{ L} = 38 \text{ mL}

2. What is the concentration of NaCl obtained?

*   **2.a. Number of moles of NaCl formed (stoichiometry):**
    *   From the equation: n_{NaCl} = 2 \times n_{Na_2CO_3}
    *   n_{NaCl} = 2 \times 0.019 \text{ mol} = 0.038 \text{ mol}
*   **2.b. Total volume of solution (V_{\text{tot}}):**
    *   V_{\text{tot}} = \text{Volume of } Na_2CO_3 \text{ solution} + \text{Volume of } HCl \text{ solution}
    *   V_{\text{tot}} = 20 \text{ mL} + 38 \text{ mL} = 58 \text{ mL} = 0.058 \text{ L}
*   **2.c. Concentration of NaCl:**
    *   C_{NaCl} = \frac{n_{NaCl}}{V_{\text{tot}}} = \frac{0.038 \text{ mol}}{0.058 \text{ L}} = 0.65 \text{ mol} \cdot \text{L}^{-1} = 0.65 \text{ M}

3. What is the volume of CO_2 released in Standard Temperature and Pressure (STP) conditions?

*   **3.a. Number of moles of CO_2 (stoichiometry):**
    *   From the equation: n_{CO_2} = n_{Na_2CO_3}
    *   n_{CO_2} = 0.019 \text{ mol}
*   **3.b. Volume of CO_2 at STP (using Ideal Gas Law):**
    *   STP conditions: T = 0^{\circ}\text{C} (273.15 \text{ K}), P = 1 \text{ atm}. Gas constant R = 0.0821 \text{ atm} \cdot \text{L} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}.
    *   V_{CO_2} = \frac{n_{CO_2} \times R \times T}{P} = \frac{0.019 \text{ mol} \times 0.0821 \text{ atm} \cdot \text{L} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \times 273.15 \text{ K}}{1 \text{ atm}}
    *   V_{CO_2} = 0.426 \text{ L} = 426 \text{ mL}

Module 8: Stoichiometry - Limiting Reagents and Yields

A. Example 3: Multi-step reaction with excess reagents

Problem: How many moles of PCl3 will be prepared from 0.2743 \text{ moles} HCl and excess MnO2 and P_4?
- Reaction a: 4HCl(aq) + MnO2(s) \rightarrow MnCl2(aq) + 2H2O(l) + Cl2(g)
- Reaction b: P4(s) + 6Cl2(g) \rightarrow 4PCl_3(s)
Steps:
1. Identify target species and knowns: Looking for moles of PCl3, known moles of HCl. Excess MnO2 and P4 mean HCl is the limiting reagent for the first step, and Cl2 for the second step.
2. Identify intermediate: Cl_2 is produced in reaction (a) and consumed in reaction (b).
3. Establish stoichiometric ratios for both reactions:
  - Reaction a: Moles of HCl to moles of Cl2: \frac{n{Cl2}}{n{HCl}} = \frac{1 \text{ mol } Cl2}{4 \text{ mol } HCl} (or n{Cl2} = \frac{1}{4} n{HCl})
  - Reaction b: Moles of Cl2 to moles of PCl3: \frac{n{PCl3}}{n{Cl2}} = \frac{4 \text{ mol } PCl3}{6 \text{ mol } Cl2} (or n{PCl3} = \frac{4}{6} n{Cl2} = \frac{2}{3} n{Cl2})
4. Combine stoichiometric ratios: Substitute n{Cl2} from (a) into (b):
  - n{PCl3} = \frac{2}{3} \times (\frac{1}{4} n{HCl}) = \frac{1}{6} n{HCl}
5. Calculate moles of PCl_3:
  - n{PCl3} = \frac{1}{6} \times 0.2743 \text{ mol HCl} = 0.04572 \text{ mol} (rounded to 5 significant figures consistent with input)

B. Limiting Reagent (or Reactant) Problems

Important Note: Unless explicitly stated that other reagents are in excess, never assume which reagent is the limiting reagent.
Techniques to identify the Limiting Reagent (LR):
1. Convert all reagent masses to moles.
2. Divide the number of moles of each reagent by its stoichiometric coefficient from the balanced equation.
3. Inspect the obtained numbers: The smallest value indicates the Limiting Reagent.
Consequence of LR: The moles (and mass) of products formed will only depend on the moles of the limiting reagent. All other reactants are in excess.

C. Example 4: Identifying the Limiting Reagent and Calculating Maximum Product Mass

Reaction: Na2CO3 + 2Na2S + 4SO2 \rightarrow 3Na2S2O3 + CO2
Problem: Calculate the maximum mass of Na2S2O_3 (MM = 158.1 \text{ g/mol}) that can be made by reacting:
- 6.00 \text{ g} of Na2CO3 (MM = 106.0 \text{ g/mol})
- 5.00 \text{ g} of Na_2S (MM = 78.0 \text{ g/mol})
- 4.00 \text{ g} of SO_2 (MM = 64.1 \text{ g/mol})
Decision Process (mass \rightarrow moles \rightarrow (moles/coefficient) \rightarrow identify LR):

Reactant	Mass (g)	MM (g/mol)	Moles (n)	Reaction Coefficient	Moles/Coefficient (n/coeff)	Limiting Reagent Status
Na2CO3	6.00	106.0	6.00 / 106.0 = 0.0566	1	0.0566 / 1 = 0.0566	Excess
Na_2S	5.00	78.0	5.00 / 78.0 = 0.0641	2	0.0641 / 2 = 0.0320	Excess
SO_2	4.00	64.1	4.00 / 64.1 = 0.0624	4	0.0624 / 4 = 0.0156	Limiting Reagent (LR)

Conclusion: SO2 has the smallest \text{moles/coefficient} value (0.0156), so SO2 is the limiting reagent.
Calculate maximum mass of Na2S2O_3 (Theoretical Yield):
- Use the moles of the limiting reagent (SO_2) and the stoichiometric ratio from the balanced equation:
  - \frac{n{Na2S2O3}}{n{SO2}} = \frac{3}{4}
  - n{Na2S2O3} = \frac{3}{4} \times n{SO2} = \frac{3}{4} \times 0.0624 \text{ mol} = 0.0468 \text{ mol}
- Convert moles of Na2S2O_3 to mass:
  - m{Na2S2O3} = n{Na2S2O3} \times MM(Na2S2O_3) = 0.0468 \text{ mol} \times 158.1 \text{ g/mol} = 7.40 \text{ g}

D. Reaction Yields - 4 Types

For a general reaction: A + B \rightarrow C (where A is the limiting reagent)

Theoretical Yield:
- The maximum mass (g) of product (C) that can be calculated based on the amount of the limiting reagent.
Actual Yield:
- The actual mass (g) of product (C) obtained when the experiment is performed.
Percentage Yield (% yield):
- Formula: \text{% yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%
- Cannot be more than 100\% (unless impurities or measurement errors are present).
Fractional Yield:
- Formula: \text{Fractional yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}
- Cannot be more than 1.

E. Example 5: Yield Calculations

Context: Using the reaction from Example 4: Na2CO3 + 2Na2S + 4SO2 \rightarrow 3Na2S2O3 + CO2
- Limiting Reagent: SO_2 (4.00 \text{ g})
- Theoretical Yield of Na2S2O_3: 7.40 \text{ g} (calculated in Example 4)
- Actual Yield of Na2S2O_3 obtained in experiment: 0.652 \text{ g}

1. Calculate Fractional Yield and Percentage Yield

*   **Fractional Yield:** \frac{\text{Actual yield}}{\text{Theoretical yield}} = \frac{0.652 \text{ g}}{7.40 \text{ g}} = 0.0881
*   **Percentage Yield:** 0.0881 \times 100\% = 8.81\%

2. Back-calculation: How much SO_2 is needed to produce a desired actual yield?

*   **Problem:** We want to make 1.20 \text{ g} of Na_2S_2O_3, assuming the 8.81\% yield obtained above. How much SO_2 should be reacted (with excess Na_2CO_3 and Na_2S)?
*   **Approach:** Use the experimental ratio of actual product obtained per reactant mass.
    *   From the experiment: 0.652 \text{ g} of Na_2S_2O_3 was obtained from 4.00 \text{ g} of SO_2 reacted.
    *   Set up a ratio to find the required SO_2 mass:
        *   \frac{\text{mass of } SO_2 \text{ needed}}{\text{desired actual mass of } Na_2S_2O_3} = \frac{\text{mass of } SO_2 \text{ reacted (exp.)}}{\text{actual mass of } Na_2S_2O_3 \text{ obtained (exp.)}}
        *   \text{mass } SO_2 = \frac{4.00 \text{ g } SO_2}{0.652 \text{ g } Na_2S_2O_3} \times 1.20 \text{ g } Na_2S_2O_3 = 7.36 \text{ g}

F. Example 6: Planning for an experiment with varying efficiency

Reaction: 4HCl(aq) + MnO2(s) \rightarrow MnCl2(aq) + Cl2(g) + 2H2O(l)
Given: We want to prepare 2.00 \text{ g} Cl2. We have 10.0 \text{ M} HCl solution and excess MnO2. MM(Cl_2) = 70.906 \text{ g} \cdot \text{mol}^{-1}.

a) If the reaction is 100\% efficient:

1.  **Moles of desired Cl_2:**
    *   n_{Cl_2} = \frac{m_{Cl_2}}{MM(Cl_2)} = \frac{2.00 \text{ g}}{70.906 \text{ g} \cdot \text{mol}^{-1}} = 0.0282 \text{ mol} (This is the theoretical yield in moles)
2.  **Moles of HCl required (stoichiometry):**
    *   From equation: n_{HCl} = 4 \times n_{Cl_2}
    *   n_{HCl} = 4 \times 0.0282 \text{ mol} = 0.113 \text{ mol}
3.  **Volume of HCl solution needed:**
    *   V_{HCl} = \frac{n_{HCl}}{C_{HCl}} = \frac{0.113 \text{ mol}}{10.0 \text{ mol} \cdot \text{L}^{-1}} = 0.0113 \text{ L} = 11.3 \text{ mL}

b) If the reaction is only 31\% efficient:

*   **Problem:** The theoretical yield of Cl_2 for 11.3 \text{ mL} of HCl is 2.00 \text{ g} (from part a). If the reaction is 31\% efficient, this means the actual yield for this amount of HCl would *not* be 2.00 \text{ g}. To ensure 2.00 \text{ g} is *actually produced*, we need to adjust the amount of HCl.
*   **Understanding the yield:**
    *   \text{% yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% = 31\%
    *   If we used 11.3 \text{ mL} of HCl, the actual yield would be:
        *   \text{Actual yield of } Cl_2 = \frac{\text{% yield}}{100} \times \text{Theoretical yield} = \frac{31}{100} \times 2.00 \text{ g} = 0.620 \text{ g} (This is not the desired 2.00 \text{ g}).
*   **Calculating adjusted HCl volume:** We can use the experimental ratio from this 31\% efficient reaction.
    *   An actual yield of 0.620 \text{ g} of Cl_2 was obtained using 11.3 \text{ mL} of HCl.
    *   To get a desired actual yield of 2.00 \text{ g} of Cl_2:
        *   \text{Volume of HCl needed} = \frac{11.3 \text{ mL HCl}}{0.620 \text{ g } Cl_2} \times 2.00 \text{ g } Cl_2 = 36.5 \text{ mL}

Module 9: Chemical Reactions - Types of Chemical Reactions and Solution Chemistry

1. Types of Chemical Reactions

a) Precipitation Reactions

Definition: Occur when ionic substances are mixed, leading to the formation of a solid ionic substance (a precipitate).
Example 1 (Precipitation): BaCl2(aq) + K2SO4(aq) \rightarrow 2KCl(aq) + BaSO4(s)
- Here, BaSO_4 is the solid precipitate.
Example 2 (No Precipitation): BaCl2(aq) + KNO3(aq) \rightarrow \text{no reaction}
- This indicates all potential products remain soluble in solution.

b) Acid-Base Reactions

Definition: Reactions between an