Stoichiometry, Limiting Reactants, and Solution Preparation in General Chemistry

Stoichiometry and Limiting Reactants

Introduction to Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction using balanced chemical equations.

• Balanced Chemical Equation: Shows the ratio in which reactants combine and products form.

• Mole Concept: The mole is a fundamental unit for counting particles in chemistry. 1 mole = $6.022 \times 10^{23}$ particles.

• Stoichiometric Coefficients: Indicate the relative number of moles of each substance involved.

Example: $2 \text{NaI} (aq) + 1 \text{Pb(NO}_3)_2 (aq) \rightarrow 2 \text{NaNO}_3 (aq) + 1 \text{PbI}_2 (s)$

Limiting Reactant and Excess Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, thus determining the maximum amount of product that can be formed. The excess reactant is the substance left over after the reaction is complete.

• Identifying the Limiting Reactant: Compare the mole ratio of reactants used to the ratio in the balanced equation.

• Calculating Product: Use the amount of limiting reactant to determine the amount of product formed.

Example Calculation:

• Given: 0.020 mol NaI and 0.010 mol Pb(NO3)2

• Reaction: $2 \text{NaI} + 1 \text{Pb(NO}_3)_2 \rightarrow 2 \text{NaNO}_3 + 1 \text{PbI}_2$

• NaI and Pb(NO3)2 are perfectly matched (no limiting reactant).

• Calculate moles of PbI2 produced: $0.010 \text{mol Pb(NO}_3)_2 \times \frac{1 \text{mol PbI}_2}{1 \text{mol Pb(NO}_3)_2} = 0.010 \text{mol PbI}_2$

• Calculate mass: $0.010 \text{mol PbI}_2 \times \frac{461.008 \text{g}}{1 \text{mol PbI}_2} = 4.61 \text{g PbI}_2$

Summary Table: Stoichiometric Calculations

Step	NaI (mol)	Pb(NO3)2 (mol)	NaNO3 (mol)	PbI2 (mol)
Initial	0.020	0.010	0	0
Change	-0.020	-0.010	+0.020	+0.010
Final	0	0	0.020	0.010

Theoretical Yield and Percent Yield

Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants, assuming complete reaction and no losses.

• Calculated using the limiting reactant.

• Expressed in grams or moles.

Formula: $\text{Theoretical Yield} = \text{Moles of Limiting Reactant} \times \frac{\text{Moles of Product}}{\text{Moles of Limiting Reactant}} \times \text{Molar Mass of Product}$

Percent Yield

The percent yield measures the efficiency of a reaction by comparing the actual yield to the theoretical yield.

• Actual Yield: The amount of product actually obtained from the reaction.

• Percent Yield Formula:

$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$ Example:

• If 10.0 g of PCl5 must be produced and the percent yield is 86%, calculate the theoretical yield and the required amount of reactants.

• Find moles of PCl5: $10.0 \text{g} \times \frac{1 \text{mol}}{208.22 \text{g}} = 0.04803 \text{mol}$

• Theoretical yield: $0.04803 \text{mol} / 0.86 = 0.05584 \text{mol}$

Solution Preparation and Volumetric Analysis

Preparing Solutions

Preparing a solution of known concentration involves dissolving a measured amount of solute in a solvent and diluting to a precise final volume.

• Volumetric Flask: Used for precise volume measurements.

• Meniscus: The curved surface of a liquid in a container; the bottom of the meniscus should be at the calibration line for accurate measurement.

• Steps:

  1. Measure out solid and dissolve in a beaker using less than the final volume.

  2. Transfer solution to volumetric flask analytically.

  3. Swirl and dilute to about 2/3 volume, swirl again.

  4. Carefully bring the volume to the line, ensuring the meniscus is just touching the line.

  5. Stopper and invert the flask to mix thoroughly.

Why is the line drawn on the neck? The narrow neck allows for precise volume measurement, minimizing error.

Molarity Calculations

Molarity (M) is defined as the number of moles of solute per liter of solution.

• Formula: $M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}$

• Example: $M = \frac{0.00181 \text{mol}}{0.116 \text{L}} = 0.0156 \text{M}$

Worked Examples and Practice Problems

Example: Limiting Reactant and Product Mass

• Given: 65 mL of 0.25 M NaI and 50 mL of 0.20 M Pb(NO3)2

• Calculate moles: $0.25 \text{M} \times 0.065 \text{L} = 0.01625 \text{mol NaI}$; $0.20 \text{M} \times 0.050 \text{L} = 0.010 \text{mol Pb(NO}_3)_2$

• Limiting reactant: Pb(NO3)2

• Mass of PbI2 formed: $0.010 \text{mol} \times 461.008 \text{g/mol} = 4.61 \text{g}$

Example: Percent Yield Calculation

• Actual yield: 10.0 g PCl5

• Theoretical yield: 11.6 g PCl5

• Percent yield: $\frac{10.0}{11.6} \times 100 = 86\%$

Practice Problem: Iron(III) Bromide Synthesis

Compound	Molar Mass (g/mol)
FeBr3	268.33
Br2	159.80
Fe2Br6	804.75

Practice Problem: Chlorine Gas Generation

• Reaction: $K_2Cr_2O_7 + 14 HCl \rightarrow 2 KCl + 2 CrCl_3 + 7 H_2O + 3 Cl_2$

• Given: 6.13 g sample of 96.0% K2Cr2O7 by mass, 320.0 mL HCl (density 1.15 g/mL, 30.0% HCl by mass)

• Calculate grams Cl2 produced at 89% efficiency

• Calculate grams of unreacted reagents

Key Terms and Definitions

• Stoichiometry: The calculation of reactants and products in chemical reactions.

• Limiting Reactant: The reactant that determines the maximum amount of product formed.

• Excess Reactant: The reactant that remains after the reaction is complete.

• Theoretical Yield: The maximum possible amount of product.

• Actual Yield: The amount of product actually obtained.

• Percent Yield: The ratio of actual yield to theoretical yield, expressed as a percentage.

• Molarity (M): Moles of solute per liter of solution.

Summary Table: Solution Preparation Steps

Step	Description
1	Measure solid and dissolve in beaker (use less than final volume)
2	Transfer solution to volumetric flask
3	Swirl and dilute to 2/3 volume, swirl again
4	Bring volume to line, meniscus at line
5	Stopper and invert flask to mix

Additional info:

• Some calculations and tables were inferred from context and standard stoichiometry procedures.

• Practice problems are typical for General Chemistry and reinforce concepts of limiting reactant, theoretical yield, percent yield, and solution preparation.