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: $H<em>2(g) + N</em>2(g) \rightarrow NH_3(g)$

  • When balanced: $3H<em>2(g) + N</em>2(g) \rightarrow 2NH_3(g)$

  • 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 $C<em>2H</em>5OH(l) + 3O<em>2(g) \rightarrow 2CO</em>2(g) + 3H<em>2O(l)$ at the molecular scale, one molecule of $C</em>2H<em>5OH$ would weigh $7.65 \times 10^{-23} \text{ g}$ and three molecules of $O</em>2$ 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.

  • $MM(C<em>2H</em>5OH) = 46.08 \text{ g/mol}$

  • $MM(O_2) = 32.00 \text{ g/mol}$

• 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:

  • $\text{gram per molecule of } C<em>2H</em>5OH = \frac{46.08 \text{ g/mol}}{6.02 \times 10^{23} \text{ molecules/mol}} = 7.65 \times 10^{-23} \text{ g/molecule}$

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 $I<em>2$ is produced if $13.1 \text{ g}$ KI is reacted with excess $KIO</em>3$ and $HNO<em>3$? (Reaction: $5KI(aq) + KIO</em>3(aq) + 6HNO<em>3(aq) \rightarrow 6KNO</em>3(aq) + 3I<em>2(aq) + 3H</em>2O(l)$)

• 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):

     • $n<em>{I</em>2 \text{ produced}} = \frac{3 \text{ moles } I<em>2}{5 \text{ moles } KI} \times n</em>{KI} = \frac{3}{5} \times 7.8916 \times 10^{-2} \text{ mol} = 4.7349 \times 10^{-2} \text{ mol } I_2$

  3. Convert moles of $I<em>2$ to mass of $I</em>2$:

     • $m<em>{I</em>2} = n<em>{I</em>2 \text{ produced}} \times MM(I_2) = 4.7349 \times 10^{-2} \text{ mol} \times 253.8 \text{ g/mol} = 12.0 \text{ g}$ (rounded to 3 significant figures)

D. Stoichiometric Calculations - Example 2

• Reaction: $Na<em>2CO</em>3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H<em>2O(l) + CO</em>2(g)$

• Reactants:

  • $Na<em>2CO</em>3$: $m = 2.0 \text{ g}$, $MM = 105.99 \text{ g} \cdot \text{mol}^{-1}$. 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 $PCl<em>3$ will be prepared from $0.2743 \text{ moles}$ HCl and excess $MnO</em>2$ and $P_4$?

  • Reaction a: $4HCl(aq) + MnO<em>2(s) \rightarrow MnCl</em>2(aq) + 2H<em>2O(l) + Cl</em>2(g)$

  • Reaction b: $P<em>4(s) + 6Cl</em>2(g) \rightarrow 4PCl_3(s)$

• Steps:

  1. Identify target species and knowns: Looking for moles of $PCl<em>3$, known moles of HCl. Excess $MnO</em>2$ and $P<em>4$ mean HCl is the limiting reagent for the first step, and $Cl</em>2$ 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 $Cl<em>2$: $\frac{n</em>{Cl<em>2}}{n</em>{HCl}} = \frac{1 \text{ mol } Cl<em>2}{4 \text{ mol } HCl}$ (or $n</em>{Cl<em>2} = \frac{1}{4} n</em>{HCl}$)

     • Reaction b: Moles of $Cl<em>2$ to moles of $PCl</em>3$: $\frac{n<em>{PCl</em>3}}{n<em>{Cl</em>2}} = \frac{4 \text{ mol } PCl<em>3}{6 \text{ mol } Cl</em>2}$ (or $n<em>{PCl</em>3} = \frac{4}{6} n<em>{Cl</em>2} = \frac{2}{3} n<em>{Cl</em>2}$)

  4. Combine stoichiometric ratios: Substitute $n<em>{Cl</em>2}$ from (a) into (b):

     • $n<em>{PCl</em>3} = \frac{2}{3} \times (\frac{1}{4} n<em>{HCl}) = \frac{1}{6} n</em>{HCl}$

  5. Calculate moles of $PCl_3$:

     • $n<em>{PCl</em>3} = \frac{1}{6} \times 0.2743 \text{ mol HCl} = 0.04572 \text{ mol}$ (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: $Na<em>2CO</em>3 + 2Na<em>2S + 4SO</em>2 \rightarrow 3Na<em>2S</em>2O<em>3 + CO</em>2$

• Problem: Calculate the maximum mass of $Na<em>2S</em>2O_3$ ($MM = 158.1 \text{ g/mol}$) that can be made by reacting:

  • $6.00 \text{ g}$ of $Na<em>2CO</em>3$ ($MM = 106.0 \text{ g/mol}$)

  • $5.00 \text{ g}$ 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):

• Conclusion: $SO<em>2$ has the smallest $\text{moles/coefficient}$ value ($0.0156$), so $SO</em>2$ is the limiting reagent.

• Calculate maximum mass of $Na<em>2S</em>2O_3$ (Theoretical Yield):

  • Use the moles of the limiting reagent ($SO_2$) and the stoichiometric ratio from the balanced equation:

    • $\frac{n<em>{Na</em>2S<em>2O</em>3}}{n<em>{SO</em>2}} = \frac{3}{4}$

    • $n<em>{Na</em>2S<em>2O</em>3} = \frac{3}{4} \times n<em>{SO</em>2} = \frac{3}{4} \times 0.0624 \text{ mol} = 0.0468 \text{ mol}$

  • Convert moles of $Na<em>2S</em>2O_3$ to mass:

    • $m<em>{Na</em>2S<em>2O</em>3} = n<em>{Na</em>2S<em>2O</em>3} \times MM(Na<em>2S</em>2O_3) = 0.0468 \text{ mol} \times 158.1 \text{ g/mol} = 7.40 \text{ g}$

D. Reaction Yields - 4 Types

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

1. Theoretical Yield:

   • The maximum mass (g) of product (C) that can be calculated based on the amount of the limiting reagent.

2. Actual Yield:

   • The actual mass (g) of product (C) obtained when the experiment is performed.

3. 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).

4. 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: $Na<em>2CO</em>3 + 2Na<em>2S + 4SO</em>2 \rightarrow 3Na<em>2S</em>2O<em>3 + CO</em>2$

  • Limiting Reagent: $SO_2$ ($4.00 \text{ g}$)

  • Theoretical Yield of $Na<em>2S</em>2O_3$: $7.40 \text{ g}$ (calculated in Example 4)

  • Actual Yield of $Na<em>2S</em>2O_3$ 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: $4HCl(aq) + MnO<em>2(s) \rightarrow MnCl</em>2(aq) + Cl<em>2(g) + 2H</em>2O(l)$

• Given: We want to prepare $2.00 \text{ g}$ $Cl<em>2$. We have $10.0 \text{ M}$ HCl solution and excess $MnO</em>2$. $MM(Cl_2) = 70.906 \text{ g} \cdot \text{mol}^{-1}$.

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): $BaCl<em>2(aq) + K</em>2SO<em>4(aq) \rightarrow 2KCl(aq) + BaSO</em>4(s)$

  • Here, $BaSO_4$ is the solid precipitate.

• Example 2 (No Precipitation): $BaCl<em>2(aq) + KNO</em>3(aq) \rightarrow \text{no reaction}$

  • This indicates all potential products remain soluble in solution.

b) Acid-Base Reactions

• Definition: Reactions between an