Stoichiometric Calculations

Learning Objective 4.5.A Explain changes in the amounts of reactants and products based on the balanced reaction equation for a chemical process.

Quick Notes

Mole ratios = coefficients. A balanced equation conserves atoms; the coefficients are exact mole ratios that link reactants and products.
Stoichiometry calculations are based on mole conversions.
- From mass: n = m / M
- Solutions: n = M × V (L)
- Gases: PV = nRT
- Particles: N = n × N_A
Limiting reagent. If two reactants are given, compute product from each; the smaller answer identifies the limit. Excess = initial − used.
Yields. Theoretical yield from stoichiometry; percent yield = (actual ÷ theoretical) × 100%.
Gas stoichiometry At the same Temperature & Pressure → volume ratios follow coefficients; otherwise use PV = nRT
Solution stoichiometry, For titrations: find moles of known concentration solution (M × V), apply the mole ratio, then solve for unknown M or V.
Units - always balance first; convert °C→K; use L and atm with the matching R; report sig figs sensibly.

Full Notes

Balanced Equations & Mole Ratios

A balanced chemical equation shows the smallest whole-number ratio of reacting species. Those coefficients are moles, so they act as exact conversion factors between substances.

Example:2 H₂ + O₂ → 2 H₂O
gives the ratios 2 mol H₂ : 1 mol O₂ : 2 mol H₂O.

Mass–Mass, Mass–Volume, and Particle Calculations

Convert to moles using molar mass (or n = M × V for solutions; or n = PV/RT for gases).
Apply the mole ratio from the balanced equation to switch substances.
Convert from moles to requested units: grams, liters (gas), liters/molarity (solution), or particles (via Avogadro’s number).

Matt’s Exam Tip

Things to look out for

Forgetting to balance the equation (wrong ratios).
Using grams directly in ratios—always convert to moles first.
Ignoring the limiting reagent when both reactants are provided.
In gas problems, mixing units (use K, atm, and the matching R; convert °C→K).
In solution problems, remember volume in liters for molarity.

Worked Example 1 — Mass → Mass

Water from Oxygen

Question: Using 2 H₂ + O₂ → 2 H₂O, what mass of water forms from 5.00 g O₂ (excess H₂)?

Moles O₂: 5.00 g ÷ 32.00 g·mol⁻¹ = 0.1563 mol.
Mole ratio to H₂O: 0.1563 mol O₂ × (2 mol H₂O / 1 mol O₂) = 0.3126 mol H₂O.
Mass H₂O: 0.3126 mol × 18.02 g·mol⁻¹ = 5.63 g.

Answer: 5.63 g H₂O.

Worked Example 2 — Limiting Reagent & Percent Yield

Ammonia Synthesis

Given: N₂ + 3 H₂ → 2 NH₃. Mix 25.0 g N₂ and 10.0 g H₂. Find the limiting reagent and theoretical mass of NH₃.

n(N₂) = 25.0 g ÷ 28.02 = 0.893 mol;
n(H₂) = 10.0 g ÷ 2.016 = 4.96 mol.
Needed H₂ for 0.893 mol N₂: 3×0.893 = 2.68 mol (available 4.96 mol) → N₂ is limiting.
n(NH₃) = 0.893 × (2/1) = 1.786 mol → mass = 1.786 × 17.03 = 30.4 g.

Theoretical yield: 30.4 g NH₃. If the actual yield were 27.5 g, percent yield = 27.5/30.4 × 100% = 90.5%.

Gas Stoichiometry

At a single temperature and pressure, gas volumes are proportional to moles, so you can use the equation’s coefficients as volume ratios. Otherwise use PV = nRT.

Same Templerature & Pressure example: For N₂ + 3 H₂ → 2 NH₃, reacting 12.0 L H₂ needs 4.0 L N₂ and makes 8.0 L NH₃.
General case: Find moles via n = PV/RT, apply the ratio, then convert back to volume if needed.

Solution Stoichiometry

Use molarity to connect volume and moles: n = M × V (V in liters). Titrations are just stoichiometry in solution.

Acid–Base Titration

Question: What is the concentration of NaOH if 25.00 mL of 0.1000 M HCl neutralizes 20.00 mL of NaOH? (Reaction: HCl + NaOH → NaCl + H₂O)

n(HCl) = 0.1000 mol·L⁻¹ × 0.02500 L = 0.002500 mol.
1:1 ratio → n(NaOH) at equivalence = 0.002500 mol.
M(NaOH) = n/V = 0.002500 ÷ 0.02000 = 0.125 M.

Answer: 0.125 M NaOH.