Stoichiometric vs fuel-rich vs fuel-lean combustion: Worked examples

See 5.3.01 Stoichiometric vs fuel-rich vs fuel-lean combustion (Hinglish) for the Hinglish parent.

The scenario matrix

Every combustion-ratio problem lives in exactly one of these cells. We will hit each one.

Cell	Class of input	What makes it distinct	Example
A	$ϕ = 1$ exactly	reference case, all fuel & O₂ consumed	Ex 1
B	$ϕ < 1$ (lean)	leftover O₂ in products	Ex 2
C	$ϕ > 1$ (rich)	leftover fuel → CO/soot; O₂ is limiting	Ex 3
D	$ϕ \to 0^{+}$ (degenerate lean)	almost no fuel — flame dies	Ex 4
E	$ϕ \to \infty$ (degenerate rich)	almost no air — cannot sustain burn	Ex 4
F	oxygenated fuel ( $C_{x} H_{y} O_{z}$ )	fuel supplies some of its own O	Ex 5
G	real-world word problem	mass flows given in kg/s, back out $ϕ$	Ex 6
H	exam twist (inverted / % excess)	tests the $ϕ$ -vs-AFR inversion trap	Ex 7
I	product-mole bookkeeping (rich split)	CO/CO₂ split when O runs out	Ex 8

Figure — Stoichiometric vs fuel-rich vs fuel-lean combustion

Figure s01 — The $ϕ$ axis map. The horizontal axis is $ϕ$ . Every example below is one point (or one limiting end) on this line. Keep glancing back at it — each example asks "where am I on this axis?"

Cell A — Stoichiometric ( $ϕ = 1$ )

Worked example Ex 1 — Propane

$C_{3} H_{8}$ burned at exactly $ϕ = 1$ Find (i) the O₂ coefficient $a$ , (ii) $AFR_{st}$ , (iii) confirm no fuel and no O₂ leftover.

Forecast: Propane has 3 carbons and 8 hydrogens — more H per molecule than methane's proportion? No: $8/3 \approx 2.7$ H per C, less than methane's $4$ . So guess $AFR_{st}$ a little below methane's $17.2$ . Write your guess down.

Balance carbon & hydrogen. $C_{3} H_{8} \to 3 C O_{2} + 4 H_{2} O$ . Why this step? Every C must land in a CO₂ (gives 3), every 2 H in one H₂O (8 H → 4 H₂O).
Balance oxygen for $a$ . Right-side O atoms $= 2 (3) + 4 = 10$ , left side $= 2 a$ , so $a = 5$ . Cross-check with formula $a = x + \frac{y}{4} = 3 + \frac{8}{4} = 3 + 2 = 5$ . ✅ Why this step? Oxygen is conserved; equate atoms and solve.
Compute AFR. $M = M_{C_{3} H_{8}} = 3 (12) + 8 (1) = 44$ g/mol. $AFR_{st} = \frac{4.76 a \cdot 28.96}{M} = \frac{4.76 \times 5 \times 28.96}{44} = \frac{689.2}{44} \approx 15.66$ Why this step? $4.76 a$ = total moles of air per mole fuel (banked constant); times $28.96$ g/mol → grams air; divide by grams fuel $M$ .

Verify: $15.66 < 17.2$ ✅ matches the forecast (less H per C ⇒ less air per kg). At $ϕ = 1$ , by definition $AFR_{actual} = AFR_{st}$ , so $ϕ = 15.66/15.66 = 1$ — exactly the reference, nothing left over. Units: $\frac{g air/mol fuel}{g fuel/mol fuel} = \frac{kg air}{kg fuel}$ ✅

Cell B — Lean ( $ϕ < 1$ )

Cell C — Rich ( $ϕ > 1$ )

Cells D & E — Degenerate extremes ( $ϕ \to 0$ and $ϕ \to \infty$ )

Worked example Ex 4 — The two limits: barely-any-fuel and barely-any-air

Forecast: As fuel $\to 0$ the mixture is nearly pure air ( $ϕ \to 0$ ); as air $\to 0$ it's nearly pure fuel ( $ϕ \to \infty$ ). Guess: both extremes fail to sustain a flame — one too dilute, one starved of oxygen.

$ϕ \to 0^{+}$ (ultra-lean). Fix methane's $AFR_{st} = 17.2$ and let $AFR_{actual} \to \infty$ (huge air, tiny fuel). Then $ϕ = 17.2/ AFR_{actual} \to 0$ . Why this step? Dividing a fixed number by an ever-larger denominator drives the quotient to 0. Physical reading: fuel molecules are so spread out that released heat can't ignite neighbours → below the lean flammability limit, flame dies.
$ϕ \to \infty$ (ultra-rich). Now $AFR_{actual} \to 0^{+}$ (almost no air), so $ϕ = 17.2/ AFR_{actual} \to \infty$ . Why this step? Dividing by a denominator shrinking toward 0 blows the quotient up. Physical reading: virtually no oxygen → nothing to oxidize → above the rich flammability limit, flame dies.

Verify: Plug two concrete numbers. $AFR = 172 \Rightarrow ϕ = 17.2/172 = 0.1$ (extreme lean). $AFR = 1.72 \Rightarrow ϕ = 17.2/1.72 = 10$ (extreme rich). Both far outside the burnable band $\approx 0.5 < ϕ < 1.7$ for methane. ✅ The endpoints of the axis in figure s01 are exactly these dead zones.

Cell F — Oxygenated fuel ( $C_{x} H_{y} O_{z}$ )

Worked example Ex 5 — Ethanol

$C_{2} H_{6} O$ : fuel carries its own oxygen Find $a$ and $AFR_{st}$ . The trap: the fuel's own O reduces the O₂ you must supply.

Forecast: Because ethanol already contains an oxygen atom, it needs less external air per molecule than a pure hydrocarbon of similar size. Guess $AFR_{st}$ notably below 15.

General oxygen balance. For $C_{x} H_{y} O_{z}$ : burn to $x C O_{2} + \frac{y}{2} H_{2} O$ . Right-side O atoms $= 2 x + \frac{y}{2}$ ; but the fuel donates $z$ of them, so the O₂ must supply the rest: $2 a = 2 x + \frac{y}{2} - z \Rightarrow a = x + \frac{y}{4} - \frac{z}{2}$ Why this step? Oxygen conservation now has a source on the left (the fuel's own O), so we subtract it — this is the only change from the hydrocarbon formula.
Plug ethanol $x = 2, y = 6, z = 1$ : $a = 2 + \frac{6}{4} - \frac{1}{2} = 2 + 1.5 - 0.5 = 3$ . Why this step? Direct substitution into the oxygenated formula.
AFR. $M = M_{C_{2} H_{6} O} = 2 (12) + 6 (1) + 16 = 46$ g/mol. $AFR_{st} = \frac{4.76 \times 3 \times 28.96}{46} = \frac{413.5}{46} \approx 8.99$ Why this step? Same air-mass machinery; only $a$ and $M$ changed.

Verify: $\approx 9.0 < 15$ ✅ matches forecast — oxygenated fuels need far less air per kg, which is exactly why flex-fuel engines re-tune the AFR for ethanol. Real-world ethanol $AFR_{st} \approx 9$ . ✅

Cell G — Real-world word problem

Cell H — Exam twist (the inversion trap)

Worked example Ex 7 — "An engine runs at 30 % excess air. Find

$ϕ$ and classify." This is the classic trap: students divide AFR the wrong way, or confuse $λ$ with $ϕ$ .

Forecast: Excess air means more air than needed ⇒ lean ⇒ $ϕ < 1$ . Guess $ϕ \approx 0.77$ .

Translate "30 % excess air" to $λ$ . 30 % excess means $λ = 1 + 0.30 = 1.30$ . Why this step? $λ =$ air supplied / air needed; "30 % extra" is $1.30 \times$ the needed air.
Invert to $ϕ$ . $ϕ = 1/ λ = 1/1.30 \approx 0.769$ . Why this step? By definition $λ = 1/ ϕ$ (parent §4), so $ϕ$ is the reciprocal — this is the step people botch by forgetting the inverse.
Classify. $ϕ = 0.769 < 1$ → fuel-lean. Why this step? The $ϕ < 1$ band is lean; excess O₂ passes through and cools the flame.

Verify: Cross-check via AFR: $AFR_{actual} = λ \cdot AFR_{st} = 1.30 AFR_{st}$ , so $ϕ = AFR_{st} / (1.30 AFR_{st}) = 1/1.30 = 0.769$ ✅ — independent of which fuel, confirming $ϕ$ is fuel-free (the whole point of the dial). ✅

Cell I — Rich product bookkeeping (where the O actually goes)

Worked example Ex 8 — Methane at

$ϕ = 1.25$ : split the carbon between CO₂ and CO When rich, oxygen is the limiting reagent. How much carbon reaches CO₂, and how much stalls at CO? Assume all H still becomes H₂O (H grabs O most eagerly).

Forecast: At $ϕ = 1.25$ we supply only $1/1.25 = 80%$ of the needed O₂. A 20 % oxygen shortfall should leave a chunk of carbon as CO. Guess: roughly a third of the C ends as CO.

O₂ actually supplied. Stoichiometric $a = 2$ mol O₂ per mol CH₄. Recall the dial $ϕ = AFR_{st} / AFR_{actual}$ ; at fixed fuel, AFR is proportional to the air (hence O₂) supplied, so $O_{2}^{actual} = O_{2}^{stoich} / ϕ = a / ϕ = 2/1.25 = 1.6$ mol. Why this step? A larger $ϕ$ means a smaller AFR means less air; dividing the stoichiometric O₂ by $ϕ$ gives the reduced supply.
Fix the hydrogen demand. $C H_{4}$ has 4 H → 2 H₂O, consuming $2 \times \frac{1}{2} = 1.0$ mol O₂. Why this step? Each H₂O needs one O atom = ½ mol O₂; two H₂O ⇒ 1.0 mol O₂ locked away first.
Oxygen left for carbon. $1.6 - 1.0 = 0.6$ mol O₂ $= 1.2$ mol O atoms for the 1 carbon. Why this step? Whatever O₂ survives after water is all that carbon can use.
Solve the CO₂/CO split. Let $p$ = mol CO₂, $q$ = mol CO, with $p + q = 1$ (one C total) and O-atom balance $2 p + q = 1.2$ . Subtract: $(2 p + q) - (p + q) = p = 0.2$ , so $p = 0.2$ , $q = 0.8$ . Why this step? Two conservation equations (carbon count, oxygen count) → two unknowns → unique split.

Verify: Products: $0.2 C O_{2} + 0.8 C O + 2 H_{2} O$ . O-atom tally $= 2 (0.2) + 0.8 + 2 (1) = 0.4 + 0.8 + 2 = 3.2$ = $2 \times 1.6$ O₂ supplied (each O₂ carries 2 O atoms, $2 \times 1.6 = 3.2$ ) ✅. Carbon tally $= 0.2 + 0.8 = 1$ ✅. So 80 % of the carbon exits as toxic CO — a vivid quantitative reason rich engines pollute. Matches forecast (well over a third). ✅

Recall Self-test: place each in the matrix

AFR = 25 for propane — which cell? ::: B (lean, since $25 > 15.66$ ). A fuel with a formula $C_{x} H_{y} O_{z}$ — which cell and which formula changes? ::: F; use $a = x + \frac{y}{4} - \frac{z}{2}$ . "40 % excess air, find φ" — which cell? ::: H; $ϕ = 1/1.40 \approx 0.714$ , lean. Almost pure fuel injected, negligible air — which cell? ::: E ( $ϕ \to \infty$ ), flame cannot sustain.

Connections

Combustion Stoichiometry & Balancing — the balancing that produced $a$ .
Limiting Reagent — why rich burning stalls carbon at CO (Ex 3, Ex 8).
Soot & Incomplete Combustion — the CO/soot products of Cell C & I.
Adiabatic Flame Temperature — why $ϕ = 1$ (Cell A) is hottest.
NOx Formation (Zeldovich) — the slightly-lean high-T corner near $ϕ \approx 0.9$ .
Rocket Propulsion — Why Engines Run Fuel-Rich — deliberate Cell C operation.
Heat of Combustion & Calorific Value — energy released per kg, tied to AFR.