Stoichiometry, Mole Concept, Limiting Reagents, and Solution Chemistry
Stoichiometry: Introduction and Mole Concept
Introduction to Stoichiometry
Reaction Stoichiometry: Involves using balanced chemical equations to determine:
The required amount of each reagent (reactant).
The amount of each product formed.
Haber Process Example: $3 H2(g) + N2(g) \rightarrow 2 NH_3(g)$ illustrates a stoichiometric ratio of 3 molecules of hydrogen to 2 molecules of ammonia.
Principle: In a balanced equation, if one quantity is known, all other quantities (mass, moles, molecules, atoms) can be determined.
The Mole Concept
Necessity: Reactions cannot be carried out at the molecular scale due to extremely small masses.
Example: For $C2H5OH(l) + 3 O2(g) \rightarrow 2 CO2(g) + 3 H_2O(l)$
At molecular scale, $1$ molecule of $C2H5OH$ would weigh 7.65 \times 10^{-23} g.
$3$ molecules of $O_2$ would weigh 15.9 \times 10^{-23} g.
Molar Masses: $MM(C2H5OH) = 46.08 \text{ g/mol}$; $MM(O_2) = 32.00 \text{ g/mol}$.
Avogadro's Number (N_A): 6.02 \times 10^{23} molecules/mol.
Calculation of mass per molecule: For $C2H5OH$, (\frac{46.08 \text{ g}}{1 \text{ mol}}) \times (\frac{1 \text{ mol}}{6.02 \times 10^{23} \text{ molecules}}) = 7.65 \times 10^{-23} \text{ g/molecule}.
The Mole as a Unit: The mole (n) is used as a standard measure.
# Moles for Solids: n = \frac{\text{mass (g)}}{\text{Molar Mass (g/mol)}}.
# Moles for Solutions: Molarity ($C{\text{solute}}$ or $M$) is defined as C{\text{solute}} = \frac{\text{Number of moles of solute (mol)}}{\text{Volume of solution (L)}}.
Therefore, n = \text{Volume used (L)} \times C_{\text{solute}} \text{ (mol/L)}.
# Moles for Gases: Using the Ideal Gas Law, $PV = nRT$.
Therefore, n = \frac{P \text{ (kPa)} \times V \text{ (L)}}{R \text{ (kPa} \cdot \text{L/mol} \cdot \text{K)} \times T \text{ (K)}}.
Constants: $V$ = Volume (L or $dm^3$);
$R = 8.314 \text{ J/(K} \cdot \text{mol)} = 8.314 \text{ kPa} \cdot \text{L/(K} \cdot \text{mol)} = 0.0821 \text{ atm} \cdot \text{L/(K} \cdot \text{mol)} = 0.0831 \text{ bar} \cdot \text{L/(K} \cdot \text{mol)}$.
$T$ = Temperature (Kelvin).
$P$ = Pressure (atm, Pa, or Torr).
STP (Standard Temperature & Pressure): $T = 0^{\circ}C (273.15K)$, $P = 1 \text{ atm (101.325 kPa)}$.
# Moles for Liquids: Using density and molar mass.
\text{mass (g)} = \text{density (g/mL)} \times \text{volume (mL)}.
Therefore, n = \frac{\text{density (g/mL)} \times \text{volume (mL)}}{\text{Molar Mass (g/mol)}}.
Example 1: Mass Calculation
Problem: What mass of $I2$ is produced if 13.1 \text{ g} $KI$ is reacted with excess $KIO3$ and $HNO3$ according to the balanced equation: $5 KI(aq) + KIO3(aq) + 6 HNO3(aq) \rightarrow 6 KNO3(aq) + 3 I2(aq) + 3 H2O(l)$.
Given: Mass of $KI = 13.1 \text{ g}$; $MM(KI) = 166.0 \text{ g/mol}$; $MM(I_2) = 253.8 \text{ g/mol}$.
Stoichiometry Ratio: From the equation, 5 moles $KI$ produce 3 moles $I_2$.
Ratio: \frac{\text{n ($I_2$) produced}}{\text{n (KI) reacted}} = \frac{3 \text{ mol}}{5 \text{ mol}}.
Steps:
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}.
Convert moles of KI to moles of $I_2$ (using stoichiometry):
n{I2 \text{ produced}} = \frac{3}{5} \times n{KI \text{ reacted}} = \frac{3}{5} \times 7.8916 \times 10^{-2} \text{ mol} = 4.7349 \times 10^{-2} \text{ mol } I2.
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}.
Example 2: Solution Stoichiometry and Gas Volume
Reaction: $Na2CO3(aq) + 2 HCl(aq) \rightarrow 2 NaCl(aq) + H2O(l) + CO2(g)$.
Given: $Na2CO3$: $m = 2.0 \text{ g}$, $MM = 105.99 \text{ g} \cdot \text{mol}^{-1}$, diluted in 20 \text{ mL} water.
$HCl$: Molarity $C_{HCl} = 1 \text{ mol} \cdot \text{L}^{-1}$.
1. Volume of HCl needed to neutralize sodium carbonate
a. Moles of $Na2CO3$: n{Na2CO_3} = \frac{2.0 \text{ g}}{105.99 \text{ g} \cdot \text{mol}^{-1}} = 0.019 \text{ mol}.
b. Moles of HCl needed (stoichiometry):
Ratio: \frac{n{HCl}}{n{Na2CO3}} = \frac{2}{1}.
n{HCl} = 2 \times n{Na2CO3} = 2 \times 0.019 \text{ mol} = 0.038 \text{ mol}.
c. Volume of HCl solution:
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. Concentration of NaCl obtained
a. Moles of NaCl (stoichiometry):
Ratio: \frac{n{NaCl}}{n{Na2CO3}} = \frac{2}{1}.
n{NaCl} = 2 \times n{Na2CO3} = 2 \times 0.019 \text{ mol} = 0.038 \text{ mol}.
b. Total volume of solution:
V{tot} = V{Na2CO3} + V_{HCl} = 20 \text{ mL} + 38 \text{ mL} = 58 \text{ mL} = 0.058 \text{ L}.
c. Concentration of NaCl:
C{NaCl} = \frac{n{NaCl}}{V_{tot}} = \frac{0.038 \text{ mol}}{0.058 \text{ L}} = 0.65 \text{ mol} \cdot \text{L}^{-1} = 0.65 \text{ M}.
3. Volume of CO2 released at STP
a. Moles of CO2 (stoichiometry):
Ratio: \frac{n{CO2}}{n{Na2CO_3}} = \frac{1}{1}.
n{CO2} = n{Na2CO_3} = 0.019 \text{ mol}.
b. Volume of CO2 at STP:
STP: $T = 273.15 \text{ K}$, $P = 1 \text{ atm}$.
Use Ideal Gas Law: $PV = nRT \Rightarrow V = \frac{nRT}{P}$.
V{CO2} = \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}} = 0.426 \text{ L} = 426 \text{ mL}.
Limiting Reagents and Yields
Example 3: Multi-step Reaction Stoichiometry
Problem: How many moles of $PCl3$ will be prepared from 0.2743 moles $HCl$ and excess $MnO2$ and $P_4$?
Reactions:
a. $4 HCl(aq) + MnO2(s) \rightarrow MnCl2(aq) + 2 H2O(l) + Cl2(g)$
b. $P4(s) + 6 Cl2(g) \rightarrow 4 PCl_3(s)$
Intermediate: $Cl_2$ is produced in reaction (a) and consumed in reaction (b).
Stoichiometric Ratios:
From (a): \frac{n{Cl2}}{n_{HCl}} = \frac{1}{4}.
From (b): \frac{n{PCl3}}{n{Cl2}} = \frac{4}{6}.
Combine Ratios:
n{PCl3} = (\frac{4}{6}) \times n{Cl2} = (\frac{4}{6}) \times (\frac{1}{4}) \times n{HCl} = \frac{1}{6} \times n{HCl}.
Calculation:
n{PCl3} = \frac{1}{6} \times 0.2743 \text{ mol HCl} = 0.04572 \text{ mol } PCl_3.
Limiting Reagent (or Reactant) Problems
Principle: Unless specified as