Summary of "Stoichiometry Tutorial: Step by Step Video + review problems explained | Crash Chemistry Academy"

Stoichiometry Tutorial (Crash Chemistry Academy)

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. The coefficients in a balanced chemical equation express fixed ratios that can refer to particle counts, mole amounts, or any proportional amounts.

Key concepts

Coefficients = ratios. Example: for the reaction 3 H2 + 1 N2 → 2 NH3 the reaction always proceeds in a 3:1:2 ratio no matter the scale (particles, moles, etc.).
Molar interpretation: coefficients can be read as moles (3 mol H2 : 1 mol N2 : 2 mol NH3).
Mole–mole conversion is the central stoichiometric step: use coefficient ratios to convert moles of one substance to moles of another.
Mass ↔ moles: since we usually measure mass, convert mass to moles (using molar mass) before applying mole–mole ratios, then convert back to mass if needed.
Other common conversions:
- Particles ↔ moles: use Avogadro’s number (6.02 × 10^23 particles = 1 mol).
- Gas volume ↔ moles (at STP): 1 mol gas = 22.4 L.

General stoichiometry methodology (step-by-step)

Write and balance the chemical equation.
Identify:
- the amount given (what you start with) and its units, and
- the amount wanted (what you must find) and its units.
Build a conversion map (chain): given → (convert to moles if needed) → mole–mole conversion using coefficients → (convert from moles if needed) → desired units.
Set up the dimensional-analysis (factor-label) chain so units cancel and only the desired unit remains.
Rules for mole–mole fractions:
- Put the substance you want on top.
- Put the substance you have on the bottom so the given unit cancels.
Use molar masses (periodic table) to convert mass ↔ moles.
Use Avogadro’s number for particle ↔ mole conversions and 22.4 L/mol for gas-volume ↔ mole at STP.
Check unit cancellation at each step. If units don’t cancel properly, the setup is wrong.
Calculator tip: once the chain is correct, multiply all numerators and divide by all denominators in a single step.

Worked examples

1) Mole–mole conversions (3 H2 + N2 → 2 NH3) - Given 7.5 mol H2 → how many mol N2? Use ratio (1 mol N2 / 3 mol H2): 7.5 × (1/3) = 2.5 mol N2.

  - Given 0.8 mol NH3 → how many mol H2?  
 Use ratio (3 mol H2 / 2 mol NH3): 0.8 × (3/2) = 1.2 mol H2.

  - Note: if another reactant isn’t mentioned assume it’s in excess.

2) Mass → mass example (same ammonia reaction) - Problem: 42.0 g N2 → mass NH3 produced? - Steps: - 42.0 g N2 × (1 mol N2 / 28.0 g N2) = 1.5 mol N2 - 1.5 mol N2 × (2 mol NH3 / 1 mol N2) = 3.0 mol NH3 - 3.0 mol NH3 × (17.0 g NH3 / 1 mol NH3) = 51 g NH3 - Result: 51 g NH3

3) Mass → mass with a different reaction (combustion / oxidizer example) - Balanced reaction used (example ratio 7:4 between NH3 and O2). - Problem: mass O2 needed for 5.95 g NH3? - Steps: convert g NH3 → mol NH3 → mol O2 (via coefficient ratio) → g O2 (via molar mass). - Result: 19.6 g O2

4) Combustion of C2H6 (mass → mass) - Problem: 37.5 g C2H6 → mass CO2 produced. - Result: 110 g CO2

5) Particles → mass example - Problem: 2.8 × 10^24 molecules of C2H6 → mass H2O produced. - Steps: molecules → mol C2H6 (divide by Avogadro’s number) → mol H2O (mole–mole ratio) → grams H2O (molar mass). - Result: 251 g H2O

Practical tips and common errors

Balance the equation first — stoichiometry depends on correct coefficients.
Always check unit cancellation at each step.
Keep track of significant figures according to the given data (good laboratory practice).
If an equation or problem omits a reactant, assume it is in excess unless told otherwise.
The mole–mole conversion is the heart of every stoichiometry problem; other conversions simply get you to/from moles.

Sources / speakers featured

Presenter / narrator: Crash Chemistry Academy instructor (sole speaker in the video).
Implicit references used in examples: periodic table (for molar masses), Avogadro’s number (6.02 × 10^23), STP gas volume (22.4 L/mol).