3.1 Stoichiometry, Chemical Reactions, and Solutions

3.1 Stoichiometry, Chemical Reactions, and Solutions

Chemistry concepts on the FE Other Disciplines exam test your ability to perform quantitative mass and mole balances, identify reaction limitations, calculate theoretical and actual product yields, and analyze solution compositions. Mastery of chemical stoichiometry and solution thermodynamics is fundamental across material science, environmental engineering, and process design.

1. Fundamentals of Stoichiometry and Mass Balances

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. All chemical calculations rest on the Law of Conservation of Mass, which dictates that atoms are neither created nor destroyed during a chemical transformation.

The Mole Concept and Molar Mass

A mole (mol) represents $6.022 \times 10^{23}$ chemical entities (Avogadro's number, $N_A$). The mass of one mole of a substance is its molar mass ($M$, measured in $\text{g/mol}$ or $\text{kg/kmol}$), which is numerically equivalent to its atomic or molecular weight found on the Periodic Table.

NCEES Formula: Mass-to-Mole Conversion

$n = \frac{m}{M}$

where:

• $n$ = quantity of substance (moles, $\text{mol}$)
• $m$ = mass of substance (grams, $\text{g}$)
• $M$ = molar mass of substance (grams per mole, $\text{g/mol}$)

Balancing Chemical Reactions

A stoichiometric equation must be balanced so that the number of atoms for each element is equal on both the reactant (left) and product (right) sides. The stoichiometric coefficients ($a, b, c, d$) define the molar ratios in which species react and form: $a A + b B \rightarrow c C + d D$

For example, in the complete combustion of methane: $\text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g)$ One mole of methane reacts with two moles of diatomic oxygen to yield one mole of carbon dioxide and two moles of water vapor.

2. Limiting Reagents, Theoretical Yield, and Percent Yield

In industrial applications and chemical processes, reactants are rarely supplied in exact stoichiometric ratios. One reactant will be depleted first, stopping the reaction.

Limiting vs. Excess Reagents

• Limiting Reagent: The reactant that is completely consumed first. It determines the maximum amount of product that can be formed.
• Excess Reagent: The reactant(s) remaining after the limiting reagent has been completely consumed.

To identify the limiting reagent:

1. Convert the given mass of each reactant into moles ($n = m / M$).
2. Divide the moles of each reactant by its respective stoichiometric coefficient ($n_i / \nu_i$).
3. The reactant with the smallest ratio ($n_i / \nu_i$) is the limiting reagent.

Theoretical and Percent Yield

• Theoretical Yield: The maximum calculated mass or moles of product formed assuming 100% conversion of the limiting reagent.
• Actual Yield: The mass or moles of product actually recovered from an experiment or industrial process (often less than theoretical yield due to side reactions, incomplete conversion, or recovery losses).

NCEES Formula: Percent Yield

$\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%$

Worked Engineering Example: Propane Combustion and Yield Calculation

Problem: A burner unit combusts $100.0\text{ g}$ of propane gas ($\text{C}_3\text{H}_8$, molar mass $= 44.10\text{ g/mol}$) with $300.0\text{ g}$ of oxygen gas ($\text{O}_2$, molar mass $= 32.00\text{ g/mol}$). The balanced reaction is: $\text{C}_3\text{H}_8(g) + 5 \text{O}_2(g) \rightarrow 3 \text{CO}_2(g) + 4 \text{H}_2\text{O}(g)$

1. Determine which reactant is the limiting reagent.
2. Calculate the theoretical yield of carbon dioxide ($\text{CO}_2$, molar mass $= 44.01\text{ g/mol}$) in grams.
3. If the combustion system operates at an actual recovery of $219.1\text{ g}$ of $\text{CO}_2$, calculate the percent yield.

Solution:

Step 1: Calculate initial moles of each reactant: $n_{\text{C}_3\text{H}_8} = \frac{100.0\text{ g}}{44.10\text{ g/mol}} = 2.2676\text{ mol}$ $n_{\text{O}_2} = \frac{300.0\text{ g}}{32.00\text{ g/mol}} = 9.3750\text{ mol}$

Step 2: Compare stoichiometric ratios ($n / \nu$):

• For $\text{C}_3\text{H}_8$: $\frac{2.2676}{1} = 2.2676\text{ mol}$
• For $\text{O}_2$: $\frac{9.3750}{5} = 1.8750\text{ mol}$

Since $1.8750 < 2.2676$, oxygen ($\text{O}_2$) is the limiting reagent.

Step 3: Calculate theoretical yield of $\text{CO}_2$: Using the mole ratio between limiting reagent $\text{O}_2$ and product $\text{CO}_2$ ($5\text{ mol O}_2 : 3\text{ mol CO}_2$): $n_{\text{CO}_2,\text{ theoretical}} = 9.3750\text{ mol O}_2 \times \left( \frac{3\text{ mol CO}_2}{5\text{ mol O}_2} \right) = 5.6250\text{ mol CO}_2$ Converting moles to mass: $m_{\text{CO}_2,\text{ theoretical}} = 5.6250\text{ mol} \times 44.01\text{ g/mol} = 247.56\text{ g}$

Step 4: Calculate percent yield: $\text{Percent Yield} = \left( \frac{219.1\text{ g}}{247.56\text{ g}} \right) \times 100\% = 88.50\%$ Conclusion: Oxygen limits product formation, yielding a theoretical maximum of $247.6\text{ g}$ of $\text{CO}_2$ and achieving an $88.5%$ operational yield.

3. Solution Concentration Metrics and Volumetric Dilutions

A solution is a homogeneous mixture consisting of a solute dissolved in a solvent. Quantitative engineering calculations require expressing concentration units based on volume, mass, or mole fractions.

Definitions of Concentration Units

1. Molarity ($M$): Moles of solute per liter of total solution ($\text{mol/L}$ or $\text{M}$). $M = \frac{n_{\text{solute}}}{V_{\text{solution (L)}}}$ Note: Molarity varies slightly with temperature due to volumetric expansion of liquids.

2. Molality ($m$): Moles of solute per kilogram of solvent ($\text{mol/kg}$ or $\text{m}$). $m = \frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}}$ Note: Molality is temperature-independent.

3. Mass Percent ($% \text{w/w}$): Mass of solute divided by total solution mass, expressed as a percentage. $\% \text{w/w} = \left( \frac{m_{\text{solute}}}{m_{\text{solute}} + m_{\text{solvent}}} \right) \times 100\%$

4. Mole Fraction ($\chi_i$): Moles of component $i$ divided by the total moles of all components in the mixture. $\chi_i = \frac{n_i}{\sum n_j}$

5. Normality ($N$): Number of equivalents of solute per liter of solution ($\text{eq/L}$ or $\text{N}$). $N = n_{eq} \times M$ where $n_{eq}$ is the number of reactive equivalents per mole of solute (e.g., $n_{eq} = 2$ for $\text{H}_2\text{SO}4$ donating $2\text{ H}^+$ ions, or $n{eq} = 2$ for $\text{Ca}^{2+}$ ions).

Volumetric Dilution Law

When diluting a concentrated stock solution with pure solvent, the total moles of solute remain constant before and after dilution ($n_1 = n_2$):

NCEES Formula: Volumetric Dilution

$M_1 V_1 = M_2 V_2 \quad \text{or} \quad C_1 V_1 = C_2 V_2$

where:

• $M_1, C_1$ = initial concentration of stock solution
• $V_1$ = volume of concentrated stock solution required
• $M_2, C_2$ = desired final concentration
• $V_2$ = final volume of diluted solution

Worked Engineering Example: Acid Stock Preparation and Dilution

Problem: An environmental testing lab requires $2.50\text{ L}$ of a $0.450\text{ M}$ sulfuric acid ($\text{H}_2\text{SO}_4$) solution. The available concentrated laboratory stock acid is $98.0%\text{ w/w}$ $\text{H}_2\text{SO}_4$ with a mass density of $\rho = 1.84\text{ g/mL}$ and a molar mass of $M = 98.08\text{ g/mol}$.

1. Calculate the molarity ($M_1$) of the concentrated stock sulfuric acid solution.
2. Determine the volume ($V_1$) of stock acid in milliliters required to prepare the diluted solution.

Solution:

Step 1: Determine molarity of concentrated stock acid ($M_1$): Consider $1.00\text{ L}$ ($1000\text{ mL}$) of stock acid solution: $m_{\text{solution}} = \rho \times V = 1.84\text{ g/mL} \times 1000\text{ mL} = 1840\text{ g}$ Mass of pure $\text{H}_2\text{SO}_4$ solute in $1.00\text{ L}$: $m_{\text{solute}} = 1840\text{ g} \times 0.980 = 1803.2\text{ g}$ Moles of $\text{H}_2\text{SO}_4$ solute: $n_{\text{solute}} = \frac{1803.2\text{ g}}{98.08\text{ g/mol}} = 18.385\text{ mol}$ Concentration of stock acid: $M_1 = \frac{18.385\text{ mol}}{1.00\text{ L}} = 18.385\text{ M}$

Step 2: Calculate required stock volume ($V_1$) using $M_1 V_1 = M_2 V_2$: $(18.385\text{ M}) \times V_1 = (0.450\text{ M}) \times (2.50\text{ L})$ $V_1 = \frac{1.125\text{ M}\cdot\text{L}}{18.385\text{ M}} = 0.06119\text{ L} = 61.19\text{ mL}$ Conclusion: Measuring $61.2\text{ mL}$ of $18.4\text{ M}$ concentrated stock acid and diluting with distilled water to a final volume of $2.50\text{ L}$ produces the desired $0.450\text{ M}$ solution.

4. Summary Table of NCEES Solution Concentration Metrics