Some Basic Concepts of Chemistry · Lesson 9 of 11

Chemical Equations, Stoichiometry, and Limiting Reagents

“Read balanced equations as quantitative relationships and calculate reactant needs, yields, and leftovers.”

Learning Objectives

• Balance equations by changing coefficients while preserving chemical formulas. • Interpret coefficients as particle, mole, and appropriate gas-volume ratios. • Convert reactant and product quantities using stoichiometric ratios. • Identify the limiting reagent from the required reaction ratio. • Calculate product amounts and excess reactant remaining.

A reaction equation is a particle account

A chemical equation names the starting substances, called reactants, and the substances formed, called products. It can also state whether each is a solid, liquid, gas, or aqueous solution. To use the equation quantitatively, we first need a correct formula for every substance and a balanced count of each element’s atoms.

The state symbols $(s)$ , $(l)$ , $(g)$ , and $(a q)$ mean solid, liquid, gas, and dissolved in water. A number written in front of a formula is a coefficient: it multiplies the entire formula. A subscript belongs to the identity of the substance itself, so it cannot be altered simply to balance an equation.

Example — Balancing a phosphorus reaction

Problem
Balance $P_{4} + O_{2} \to P_{4} O_{10}$ .

1.The phosphorus count is already four on both sides.
2.The product contains ten O atoms. Each reactant $O_{2}$ supplies two, so five oxygen molecules are required.
3.The balanced equation is $P_{4} + 5 O_{2} \to P_{4} O_{10}$ . Changing the product subscript to 2 would instead change the stated compound.

Other examples are $4Fe + 3 O_{2} \to 2F e_{2} O_{3}$ and $2Mg + O_{2} \to 2MgO$ . Count atoms rather than the number of written formulas. Two formula units of $F e_{2} O_{3}$ contain four iron atoms and six oxygen atoms.

Example — Stepwise combustion balancing

Problem
Balance the complete combustion of propane, $C_{3} H_{8}$ , to carbon dioxide and water.

1.Start with $C_{3} H_{8} + O_{2} \to C O_{2} + H_{2} O$ . These are correct formulas; coefficients are still to be found.
2.Balance C: three carbon atoms require $3 C O_{2}$ . Balance H: eight hydrogens require $4 H_{2} O$ .
3.The products now contain $3 (2) + 4 (1) = 10$ oxygen atoms, requiring $5 O_{2}$ .
4.The result is $C_{3} H_{8} + 5 O_{2} \to 3C O_{2} + 4 H_{2} O$ . Check C = 3, H = 8, and O = 10 on each side. State symbols can then specify the experimental conditions.

One equation, several quantitative readings

Stoichiometry is the quantitative calculation of substances consumed or formed in a reaction. Its ratios come from a balanced equation. Because a mole is a fixed number of entities, the same coefficient ratio applies to particles and moles.

Methane combustionLaTeX

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g)

The coefficient ratio is

1 : 2 : 1 : 2

. Water is specified as vapour in this equation.

Interpretation	Methane	Oxygen	Carbon dioxide	Water vapour
Molecules	1	2	1	2
Moles	1	2	1	2
Mass, using rounded molar masses	16 g	64 g	44 g	36 g
Gas volumes at common temperature and pressure	1 volume	2 volumes	1 volume	2 volumes

The mass row is not 1:2:1:2 because the substances have different molar masses. Its total reactant mass is 80 g and total product mass is also 80 g. The volume row applies when all compared substances are gases at the same conditions; condensing the water changes that volume comparison.

Counting a balanced reaction— Each completed batch uses two hydrogen molecules and one oxygen molecule. All atoms are conserved.

A commonly used ideal-gas molar volume is about $22.7 L mo l^{- 1}$ at $273.15 K$ and $1 bar$ . At $273.15 K$ and $1 atm$ it is about $22.4 L mo l^{- 1}$ . Always specify temperature and pressure rather than treating a quoted standard volume as universal. To compare volumes of water vapour, choose conditions at which it remains gaseous; do not assume the vapour stays gaseous at a low reference temperature.

The mass-to-mole-to-mass route

Most reaction problems provide a mass but require a quantity of another substance. Convert the given mass to moles, apply the coefficient ratio, and then convert the required amount to the requested unit. A mass ratio copied directly from coefficients skips the molar-mass step and is generally wrong.

Stoichiometric amount conversionLaTeX

n_{B} = n_{A} \frac{ν _{B}}{ν _{A}}

n_{A}

and

n_{B}

are amounts of substances A and B. The symbols

ν_{A}

and

ν_{B}

are their positive coefficients in the balanced equation.

Example — Water from methane

Problem
Find the water mass formed by complete combustion of $16.0 g$ methane with excess oxygen. Use $M (C H_{4}) = 16.0$ and $M (H_{2} O) = 18.0 g mo l^{- 1}$ .

1.Methane amount is $16.0/16.0 = 1.00 mol$ .
2.The balanced equation gives two moles water for every mole methane: $n (H_{2} O) = 2 (1.00) = 2.00 mol$ .
3.Water mass is $2.00 (18.0) = 36.0 g$ . The additional mass came from oxygen, so a product mass larger than the methane mass is reasonable.

Example — Working backward from a product

Problem
How many moles of methane are needed to produce $22.0 g$ carbon dioxide? Use $M (C O_{2}) = 44.0 g mo l^{- 1}$ .

1.Convert the carbon dioxide mass: $n = 22.0/44.0 = 0.500 mol$ .
2.The methane:carbon dioxide mole ratio is 1:1, so methane required is $0.500 mol$ .
3.If methane mass were requested, multiply by its molar mass. The 1:1 relation is between amounts, not between gram values.

When one reactant runs out first

A balanced equation specifies the required ratio, but available samples may not be supplied in that ratio. The reactant that is exhausted first prevents further completion of the reaction and limits the amount of product. The other reactant can remain in excess.

Definition

Limiting reagent

The reactant that is consumed first in the stoichiometric reaction and therefore limits the amount of product that can form.

Availability must be compared with the required ratio— Six hydrogen molecules and two oxygen molecules can form only four water molecules; two hydrogen molecules remain.

For a balanced reaction, divide each reactant amount by its coefficient. This finds the number of reaction batches that each amount could support. The smallest value determines the limiting reagent. Equal values indicate an exactly stoichiometric supply, so neither reactant is left in excess under the assumed complete reaction.

Reaction capacityLaTeX

ξ_{m a x} = i min (\frac{n _{i}}{ν _{i}})

n_{i}

is the initial amount of reactant

i

and

ν_{i}

its coefficient. The maximum reaction extent

ξ_{m a x}

is a convenient measure of reaction batches in moles; product amount is its coefficient times this value.

Least mass does not identify the limiting reagent

Reactants with different molar masses and required coefficients cannot be compared by grams alone. Convert to moles and compare the amounts relative to the balanced coefficients. The smaller number of moles alone is not sufficient either.

Example — Ammonia from two available reactants

Problem
$50.0 kg$ nitrogen and $10.0 kg$ hydrogen react to form ammonia. Find the limiting reagent and ammonia mass. Use molar masses $28.0$ , $2.016$ , and $17.0 g mo l^{- 1}$ for $N_{2}$ , $H_{2}$ , and $N H_{3}$ .

1.First balance: $N_{2} + 3 H_{2} \to 2N H_{3}$ .
2.Convert to grams and moles: $n (N_{2}) = 50000/28.0 = 1785.71 mol$ ; $n (H_{2}) = 10000/2.016 = 4960.32 mol$ .
3.Compare capacities: nitrogen gives $1785.71/1 = 1785.71$ ; hydrogen gives $4960.32/3 = 1653.44$ . Hydrogen is limiting.
4.Ammonia amount is $2 (1653.44) = 3306.88 mol$ . Its mass is $3306.88 (17.0) = 56216.9 g \approx 56.2 kg$ to three significant figures.
5.Nitrogen consumed is $1653.44 mol$ , leaving $(1785.71 - 1653.44) (28.0) = 3703.7 g$ , about $3.70 kg$ . These rounded molar masses give slight mass-bookkeeping differences; internally consistent more precise masses remove that numerical artefact.

The example uses rounded molar masses to match the level of the chapter. Its calculation retains working digits until the final result. A slightly different printed yield can result from rounding each intermediate amount, which is a reason to show the method and state the molar masses rather than memorise an answer.

Example — A smaller amount need not be limiting

Problem
For $N_{2} + 3 H_{2} \to 2N H_{3}$ , compare $2.00 mol$ nitrogen and $3.00 mol$ hydrogen.

1.Although nitrogen has fewer moles, it supports $2.00/1 = 2.00 mol$ of reaction batches.
2.Hydrogen supports only $3.00/3 = 1.00 mol$ of batches, so hydrogen is limiting.
3.The products contain $2.00 mol$ ammonia; $1.00 mol$ nitrogen remains. The coefficient comparison decides the result.

Quick check

What may be changed while balancing a chemical equation?

Quick check

In $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ , the methane:oxygen ratio 1:2 is directly a ratio of:

Quick check

For $N_{2} + 3 H_{2} \to 2N H_{3}$ , $1 mol$ nitrogen and $2 mol$ hydrogen are available. Which limits the yield?

Quick check

How many moles water form from $0.500 mol$ methane with excess oxygen?

Quick check

When can gas-volume ratios be read directly from coefficients?

Practice Problems

Balance $Fe + O_{2} \to F e_{2} O_{3}$ and explain each coefficient.
Calculate carbon dioxide from $1.00 mol$ carbon with excess oxygen using $C + O_{2} \to C O_{2}$ .
Repeat for $1.00 mol$ carbon with $16.0 g$ oxygen using $M (O_{2}) = 32.0$ .
For $A + B_{2} \to A B_{2}$ , compare 300 A atoms and 200 $B_{2}$ molecules; state the remaining entities.
Calculate the ammonia and leftover reactant from $2.00 \times 1 0^{3} g$ nitrogen and $1.00 \times 1 0^{3} g$ hydrogen using molar masses 28.0, 2.00, and 17.0.
Find the HCl mass required for $5.0 g$ $Mn O_{2}$ using $4HCl + Mn O_{2} \to 2 H_{2} O + MnC l_{2} + C l_{2}$ , with molar masses 36.5 and 87.0.
Explain why a quoted molar gas volume is incomplete without temperature and pressure.

Key Takeaways

Balance atoms by coefficients while keeping formulas unchanged. Coefficients give particle and mole ratios; gas volume ratios also need common conditions. Convert mass to moles before applying a stoichiometric ratio. The smallest reactant amount divided by its coefficient identifies the limiting supply. Product yield and excess remaining follow the amount of reaction that can actually occur.

Previous · Lesson 8

Percentage Composition and Chemical Formulas

Next · Lesson 10

Expressing Concentration in Solutions

A reaction equation is a particle account

One equation, several quantitative readings

The mass-to-mole-to-mass route

When one reactant runs out first

Quiz

Practice Problems

Key Takeaways