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General Chemistry I · Chemical Process

Chemical Equations and Reaction Stoichiometry

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Want it in plain words first? Jump to Eli explains — the same idea, no jargon.
On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

A balanced shows the mole ratios (coefficients) that relate reactants and products. uses these ratios to convert between amounts of substances, typically along the path mass → moles → moles → mass. The is the one that runs out first and determines the ; the is left over. compares the to the theoretical yield.

Why this matters

Stoichiometry underlies pharmaceutical synthesis: chemists calculate the exact masses of starting materials needed to make a target amount of a drug, and percent yield tells them how efficient the synthesis was. In clinical chemistry, balanced equations and mole ratios calibrate assays (how much reagent reacts with a known amount of analyte). Limiting-reactant thinking also models biological pathways where a scarce nutrient limits formation.

The college version

1. Writing and Balancing Equations

  • Write correct formulas (never change subscripts) with reactants → products.
  • Balance by adjusting coefficients only, so atoms of each element are equal on both sides.
  • Include state symbols when useful: (s), (l), (g), (aq).
  • Example: CH₄ + 2 O₂ → CO₂ + 2 H₂O (1 C, 4 H, and 4 O on each side).
  • Tips: balance H and O last; use fractions if needed, then clear them (e.g., ½ O₂ → multiply through by 2).

2. Stoichiometric Calculations

  • Coefficients give mole ratios (stoichiometric factors).
  • Path: mass (g) → moles (÷ molar mass) → moles (× ) → mass (× molar mass).
  • Mole-to-mole: use the ratio directly.
  • Mass-to-mass: convert grams of A to moles, apply the ratio, convert moles of B to grams.

3. Limiting Reactant and Yield

  • Limiting reactant: completely consumed; determines the amount of product.
  • Excess reactant: left over after the reaction.
  • Theoretical yield: maximum product (g) from the limiting reactant.
  • Actual yield: amount actually obtained in the experiment.
  • Percent yield = (actual / theoretical) × 100%.

How it works

  1. Write correct formulas and balance the equation (coefficients only).
  2. Convert the given mass to moles using molar mass.
  3. Apply the mole ratio (coefficients) to convert moles of the known to moles of the unknown.
  4. Convert moles of the unknown back to mass (if mass is requested).
  5. For a limiting reactant, compute product moles from each ; the smaller value identifies the limiting reactant and the theoretical yield.
  6. Compute percent yield = actual ÷ theoretical × 100%.

Common confusions

Do not confuseWithDifference
CoefficientSubscriptCoefficient balances atoms; subscript is fixed identity
Mole ratioMass ratioRatios apply to moles, not grams
Limiting reactantExcess reactantLimiting is consumed; excess remains
Theoretical yieldActual yieldTheoretical = calculated max; actual = measured
Percent yieldPercent compositionYield = reaction efficiency; composition = mass % of elements

Memory aids

"Grams → Moles → Moles → Grams" (GMMG) — the mass-to-mass path: divide by molar mass, multiply by the mole ratio, multiply by molar mass.

Quick review

Topic Recap

Balanced equations give mole ratios that drive stoichiometric calculations (mass → moles → moles → mass). The limiting reactant determines the theoretical yield; the excess reactant remains. Percent yield compares actual to theoretical, revealing reaction efficiency.

Knowledge Check

  1. Balance: Fe + O₂ → __ Fe₂O₃.
  2. In 2 H₂ + O₂ → 2 H₂O, how many moles of H₂O form from 3.0 mol H₂?
  3. In N₂ + 3 H₂ → 2 NH₃, if you have 1.0 mol N₂ and 2.0 mol H₂, which is limiting?
  4. If the theoretical yield is 20.0 g and you collect 15.0 g, what is the percent yield?
  5. What mass of CO₂ forms from 12.0 g of C in C + O₂ → CO₂? (C = 12.01, CO₂ = 44.01 g/mol)

Answers and Rationales

  1. 4 Fe + 3 O₂ → 2 Fe₂O₃.
  2. 3.0 mol — 3.0 mol H₂ × (2 H₂O / 2 H₂) = 3.0 mol H₂O.
  3. H₂ — H₂ yields 2.0 × (2/3) = 1.33 mol NH₃; N₂ yields 2.0 mol; H₂ produces less.
  4. 75.0% — (15.0 / 20.0) × 100%.
  5. 44.0 g — 12.0 g C × (1/12.01) = 0.999 mol C → 0.999 mol CO₂ × 44.01 = 44.0 g CO₂.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of a balanced equation as a hamburger recipe: 2 buns + 1 patty → 1 burger. The numbers give the ratio—you need twice as many buns as patties. If you have 10 buns and 8 patties, you can make only 5 burgers because the buns run out first; buns are the "limiting reactant" and patties are "excess."

Stoichiometry is just scaling that recipe: if the recipe makes 1 burger and you want 5, you need 10 buns and 5 patties. Chemistry works the same way, except the "recipe" (the balanced equation) gives mole ratios, and we convert grams to moles (and back) to measure amounts in the lab.

Where it stops being exact: real reactions don't always go to 100% completion—some reactant is lost, or side reactions occur—so the "actual yield" is usually less than the recipe predicts. The recipe tells you the maximum possible amount (the theoretical yield), not what you'll actually collect.

Simple Example

2 H₂ + O₂ → 2 H₂O. If you start with 4 mol H₂ and 1 mol O₂, oxygen is the limiting reactant: 1 mol O₂ makes 2 mol H₂O, using only 2 mol H₂ and leaving 2 mol H₂ in excess.

Worked example

Example 1 (mole-to-mole). How many moles of O₂ react with 3.0 mol of C₃H₈ in C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O? 3.0 mol C3H8 × 5 mol O21 mol C3H8 = 15 mol O2

Example 2 (mass-to-mass). What mass of H₂O forms from 4.04 g of H₂ (2 H₂ + O₂ → 2 H₂O)? (Molar masses: H₂ = 2.016 g/mol, H₂O = 18.02 g/mol.) 4.04 g H2 × 1 mol H22.016 g H2 × 2 mol H2O2 mol H2 × 18.02 g H2O1 mol H2O = 36.1 g H2O

Example 3 (limiting reactant and yield). N₂ + 3 H₂ → 2 NH₃. Starting with 28.0 g N₂ and 4.0 g H₂, find the limiting reactant, theoretical yield of NH₃, and percent yield if 18.0 g NH₃ are actually produced. (Molar masses: N₂ = 28.02, H₂ = 2.016, NH₃ = 17.03 g/mol.)

Step 1 (moles of each reactant): N2: 28.028.02=1.00 mol; H2: 4.02.016=2.0 mol Step 2 (moles of NH₃ each could produce): From N2: 1.00 mol N2 × 2 mol NH31 mol N2 = 2.00 mol NH3 From H2: 2.0 mol H2 × 2 mol NH33 mol H2 = 1.33 mol NH3 Step 3: H₂ produces less NH₃, so H₂ is the limiting reactant and N₂ is in excess. Step 4 (theoretical yield): 1.33 mol NH3 × 17.03 g/mol = 22.7 g NH3 Step 5 (percent yield): 18.0 g22.7 g × 100% = 79.3%

Common errors: using the wrong mole ratio (inverting it); comparing grams of reactants directly instead of moles; using the excess reactant to compute theoretical yield; and confusing theoretical (calculated) with actual (measured) yield.

Key takeaways

  • High yield: Balance by changing coefficients only—never subscripts.
  • High yield: Coefficients give mole ratios, not mass ratios.
  • High yield: The limiting reactant produces the least product and determines the theoretical yield.
  • High yield: Percent yield = (actual / theoretical) × 100%, usually < 100%.
  • Always convert grams → moles before using mole ratios.
  • Compare moles of product (not grams of reactant) to find the limiting reactant.

Keep learning

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Practice General Chemistry I

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Write and balance chemical equations using correct formulas and coefficients.
  • Perform mole-to-mole and mass-to-mass stoichiometric calculations.
  • Identify the limiting and excess reactants in a reaction.
  • Calculate theoretical yield, actual yield, and percent yield.

Key vocabulary

Chemical equation
Symbols describing a reaction
Coefficient
Number before a formula
Subscript
Number within a formula
Reactant
Starting substance
Product
Substance formed
Mole ratio
Ratio of coefficients
Stoichiometry
Quantitative relationships
Limiting reactant
Runs out first
Excess reactant
Left over
Theoretical yield
Maximum product (calculated)
Actual yield
Product obtained (measured)
Percent yield
actual/theoretical × 100%

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