Chemical Reactions and Stoichiometry
Key Takeaways
- Stoichiometry uses mole ratios from balanced equations to relate quantities of reactants and products.
- The mole concept: 1 mole = 6.022 × 10²³ particles (Avogadro's number); molecular weight in g/mol.
- Chemical equations must be balanced: atoms of each element must be equal on both sides.
- Limiting reagent is the reactant that is completely consumed first, determining the maximum product yield.
- Theoretical yield is the maximum product from stoichiometry; percent yield = (actual/theoretical) × 100%.
- Chemical equilibrium is described by the equilibrium constant K = [products]/[reactants] with stoichiometric exponents.
Chemical Reactions and Stoichiometry
FE Exam Weight: Chemistry accounts for 5-8 questions (~6% of the exam). Focus on stoichiometry, acid-base chemistry, and oxidation-reduction reactions.
The Mole Concept
| Quantity | Value |
|---|---|
| Avogadro's Number | 6.022 × 10²³ particles/mol |
| Molar Mass | Numerically equal to atomic/molecular weight in g/mol |
| STP (Standard Temp & Pressure) | 0°C (273.15 K) and 1 atm |
| Molar Volume at STP | 22.4 L/mol for ideal gas |
Converting between mass, moles, and molecules:
Balancing Chemical Equations
Every chemical equation must conserve mass — the number of atoms of each element must be equal on both sides.
Example: Balance Fe₂O₃ + CO → Fe + CO₂
Step 1: Fe: 2 on left → need 2 on right → 2Fe Step 2: O: 3 + 1 = 4 on left → need 4 on right → need adjustment Step 3: Fe₂O₃ + 3CO → 2Fe + 3CO₂ (Check: Fe: 2=2, C: 3=3, O: 3+3=6 and 6=6 ✓)
Stoichiometry Calculations
Stoichiometry uses mole ratios from balanced equations to calculate amounts of reactants and products.
Steps:
- Write and balance the chemical equation
- Convert known quantity to moles
- Use mole ratio to find moles of desired substance
- Convert moles to desired units (grams, liters, etc.)
Example: How many grams of CO₂ are produced when 100 g of propane (C₃H₈) burns completely?
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
- Moles of C₃H₈ = 100 g / 44.1 g/mol = 2.268 mol
- Mole ratio: 3 mol CO₂ per 1 mol C₃H₈
- Moles of CO₂ = 2.268 × 3 = 6.804 mol
- Mass of CO₂ = 6.804 × 44.0 g/mol = 299.4 g
Limiting Reagent
The limiting reagent is the reactant that is completely consumed first, limiting the amount of product formed.
Method: Calculate the moles of product each reactant can produce. The reactant that produces the least product is the limiting reagent.
Chemical Equilibrium
For a reversible reaction: aA + bB ⇌ cC + dD
| K Value | Meaning |
|---|---|
| K >> 1 | Products are favored at equilibrium |
| K << 1 | Reactants are favored at equilibrium |
| K ≈ 1 | Neither strongly favored |
Le Chatelier's Principle
If a system at equilibrium is disturbed, it shifts to counteract the change:
| Change | Equilibrium Shifts |
|---|---|
| Add reactant | Toward products (right) |
| Remove product | Toward products (right) |
| Increase temperature (exothermic) | Toward reactants (left) |
| Increase temperature (endothermic) | Toward products (right) |
| Increase pressure | Toward fewer moles of gas |
How many moles of oxygen gas (O₂) are needed to completely combust 2 moles of methane (CH₄)? Reaction: CH₄ + 2O₂ → CO₂ + 2H₂O
For the equilibrium reaction N₂ + 3H₂ ⇌ 2NH₃ (exothermic), what happens if the temperature is increased?
In a reaction, 10 g of hydrogen (H₂, MW = 2) reacts with 80 g of oxygen (O₂, MW = 32) to form water. Which is the limiting reagent?