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Stoichiometry of the Reaction

General Chemistry • Chemical Reactions

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Reaction quantities

Stoichiometry of the Reaction

Stoichiometry uses a balanced chemical equation to predict how much reactant is needed and how much product can form.

2H2 + O2 → 2H2O

Learning target: convert between mass, moles, particles, and product yield using mole ratios from the balanced equation.

Core skill

Use coefficients

Stoichiometric ratios come from the balanced equation, not from subscripts alone.

Prediction

Find amounts

Calculate required reactant, limiting reactant, and theoretical yield.

Method

Convert through moles

Moles are the bridge between a balanced equation and measurable mass.

Why it matters

Stoichiometry turns a reaction into a quantitative recipe

A balanced chemical equation gives the particle ratio and the mole ratio. This lets chemists scale reactions from invisible particles to grams measured on a balance.

N2 + 3H2 → 2NH3

The coefficients mean \(1\ \mathrm{mol}\) N2 reacts with \(3\ \mathrm{mol}\) H2 to form \(2\ \mathrm{mol}\) NH3.

Stoichiometry helps answer

  • How many grams of product can form?
  • Which reactant runs out first?
  • How much excess reactant remains?
  • How efficient was the real experiment?

Core concept

The balanced equation is the mole-ratio map

For the formation of water, the balanced equation is:

2H2 + O2 → 2H2O

This gives the mole relationships:

  • \(2\ \mathrm{mol}\) H2 react with \(1\ \mathrm{mol}\) O2.
  • \(2\ \mathrm{mol}\) H2 form \(2\ \mathrm{mol}\) H2O.
  • \(1\ \mathrm{mol}\) O2 forms \(2\ \mathrm{mol}\) H2O.
Mole ratio map for water formation A diagram showing two moles of hydrogen gas and one mole of oxygen gas producing two moles of water. 2 mol H2 reactant amount 1 mol O2 reactant amount 2 mol H2O product amount Coefficients become mole ratios 2 : 1 : 2 is the recipe for H2, O2, and H2O

Vocabulary

Variables and quantities used in stoichiometry

Quantity Meaning Useful relationship Common unit
Mole ratio Ratio from coefficients in a balanced equation \(\frac{\mathrm{mol\ wanted}}{\mathrm{mol\ given}}\) mol/mol
Molar mass Mass of \(1\ \mathrm{mol}\) of a substance \(n = \frac{m}{M}\) g/mol
Avogadro's number Particles in \(1\ \mathrm{mol}\) \(6.022 \times 10^{23}\ \mathrm{particles/mol}\) particles/mol
Limiting reactant Reactant that runs out first Produces the smaller product amount mol or g
Theoretical yield Maximum product predicted by stoichiometry Based on the limiting reactant g or mol

Calculation pathway

Most stoichiometry problems pass through moles

Mass, particles, and gas volume are measured in different ways, but the balanced equation compares substances in moles.

Given amount mass, particles, or gas volume
Moles of given \(n = \frac{m}{M}\)
Mole ratio from coefficients
Moles wanted predicted by equation
Final amount grams, particles, or volume
\[ \mathrm{g\ given} \rightarrow \mathrm{mol\ given} \rightarrow \mathrm{mol\ wanted} \rightarrow \mathrm{g\ wanted} \]

The mole ratio step is the heart of stoichiometry.

Interactive simulation

Find the limiting reactant and theoretical yield

Reaction model

2H2 + O2 → 2H2O

Calculated prediction

mol H2 4.96 mol
mol O2 1.56 mol
Limiting reactant O2
Theoretical H2O 56.3 g

O2 is limiting because it produces less H2O.

Static fallback: with 10.0 g H2 and 50.0 g O2, O2 limits the reaction and the predicted H2O yield is 56.3 g.

Dynamic model

The smaller product prediction controls the yield

Each reactant can independently predict an amount of H2O. The lower prediction is the theoretical yield because that reactant runs out first.

Theoretical yield comparison graph A bar graph comparing water predicted from hydrogen and oxygen. 120 g 90 g 60 g 30 g 0 g From H2 From O2 89.4 g 56.3 g theoretical yield

Rule: theoretical yield is determined by the reactant that gives the smaller amount of product.

Worked example

Mass-to-mass stoichiometry with a limiting reactant

Problem: 10.0 g H2 reacts with 50.0 g O2. What is the theoretical yield of H2O?

2H2 + O2 → 2H2O
  1. 1. Convert each reactant to moles. \(n_{\mathrm{H_2}} = \frac{10.0\ \mathrm{g}}{2.016\ \mathrm{g/mol}} = 4.96\ \mathrm{mol\ H_2}\) \(n_{\mathrm{O_2}} = \frac{50.0\ \mathrm{g}}{32.00\ \mathrm{g/mol}} = 1.56\ \mathrm{mol\ O_2}\)
  2. 2. Predict product from each reactant. \(4.96\ \mathrm{mol\ H_2} \times \frac{2\ \mathrm{mol\ H_2O}}{2\ \mathrm{mol\ H_2}} = 4.96\ \mathrm{mol\ H_2O}\) \(1.56\ \mathrm{mol\ O_2} \times \frac{2\ \mathrm{mol\ H_2O}}{1\ \mathrm{mol\ O_2}} = 3.12\ \mathrm{mol\ H_2O}\)
  3. 3. Choose the smaller product amount. O2 is limiting because it produces only \(3.12\ \mathrm{mol}\) H2O.
  4. 4. Convert moles of product to grams. \(3.12\ \mathrm{mol\ H_2O} \times 18.015\ \mathrm{g/mol} = 56.3\ \mathrm{g\ H_2O}\)

Percent yield

Real experiments often produce less than the theoretical yield

Theoretical yield is the maximum amount predicted by the balanced equation. Actual yield is the amount collected in the lab.

\[ \%\,\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \]

If the theoretical yield is 56.3 g H2O and the actual yield is 48.0 g H2O:

\[ \%\,\text{yield} = \frac{48.0\ \mathrm{g}}{56.3\ \mathrm{g}} \times 100\% = 85.3\% \]

Why actual yield may be lower

  • Some product is lost during transfer.
  • The reaction may not go to completion.
  • Side reactions may form other products.
  • Measurements may contain experimental error.

Common mistake

Do not take stoichiometric ratios from subscripts alone

Incorrect reasoning

“H2O has two H atoms and one O atom, so the ratio must always be 2:1.”

This ignores the coefficients in the full balanced equation.

Correct reasoning

Use the balanced equation:

2H2 + O2 → 2H2O

The mole ratio H2:O2:H2O is 2:1:2.

Key distinction: subscripts describe the composition of one formula unit or molecule. Coefficients describe the reacting amounts in the balanced equation.

Practice check

Predict product from a balanced equation

Ammonia forms according to:

N2 + 3H2 → 2NH3

If \(6.00\ \mathrm{mol}\) H2 reacts with excess N2, how many moles of NH3 can form?

Show answer and reasoning

Use the mole ratio from the coefficients: \(3\ \mathrm{mol}\) H2 forms \(2\ \mathrm{mol}\) NH3.

\[ 6.00\ \mathrm{mol\ H_2} \times \frac{2\ \mathrm{mol\ NH_3}}{3\ \mathrm{mol\ H_2}} = 4.00\ \mathrm{mol\ NH_3} \]

Answer: \(4.00\ \mathrm{mol}\) NH3.

Apply the topic

A reliable stoichiometry strategy

Step 1

Balance the equation

Stoichiometry cannot start until the reaction coefficients are correct.

Step 2

Convert to moles

Use molar mass, Avogadro's number, or gas volume when appropriate.

Step 3

Use the mole ratio

Choose the ratio that connects the given substance to the wanted substance.

Step 4

Convert to the requested unit

Report the answer with units and reasonable significant figures.

For limiting reactant problems, calculate product from each reactant separately. The smaller product amount determines the theoretical yield.

Summary

What to remember

Coefficients give mole ratios

The balanced equation is the source of every stoichiometric ratio.

Moles are the bridge

Convert mass, particles, or volume into moles before using the equation ratio.

Limiting reactant controls yield

The reactant that produces the smaller amount of product runs out first.

Percent yield compares lab to theory

\(\%\,\text{yield} = \frac{\text{actual}}{\text{theoretical}} \times 100\%\)

2H2 + O2 → 2H2O