Combustion stoichiometry — fuel + O₂ → CO₂ + H₂O: Exercises

Before anything else, recall the quantities we'll lean on. Every symbol here was earned in the parent note:

Definition The three counters (inputs) and the three coefficients (outputs)

Counters describe the fuel molecule:

$x$ ::: number of carbon atoms in one fuel molecule (they all become $CO_{2}$ ).
$y$ ::: number of hydrogen atoms in one fuel molecule (they all become $H_{2} O$ ).
$z$ ::: number of oxygen atoms already inside the fuel (a "credit" that reduces the $O_{2}$ you must supply).

Coefficients are the big numbers we place in front of each species when balancing (from the parent's derivation):

$a$ ::: the number in front of $O_{2}$ — how many $O_{2}$ molecules the fuel needs $= x + \frac{y}{4} - \frac{z}{2}$ .
$b$ ::: the number in front of $CO_{2}$ — how many $CO_{2}$ are made $= x$ .
$c$ ::: the number in front of $H_{2} O$ — how many waters are made $= \frac{y}{2}$ .

A mole is just a fixed count of molecules ( $6.022 \times 1 0^{23}$ of them) — think of it as a "dozen" but huge. See Mole Concept and Molar Mass. The molar mass $M (species)$ is the mass in grams of one mole of that species; e.g. $M (O_{2}) = 32$ g/mol means one mole of oxygen gas weighs 32 g. This is exactly the molar-mass idea from the parent chain.

Level 1 — Recognition

You are only asked to read off or match — no heavy algebra yet.

Recall Solution L1.1

WHAT we do: just look at the subscripts.

$x = 3$ (three carbons), $y = 8$ (eight hydrogens), $z = 0$ (no oxygen written, so none inside).
$CO_{2}$ count $= b = x = 3$ . WHY $b = x$ : every carbon must land in exactly one $CO_{2}$ , and no carbon can be created or destroyed, so the count of $CO_{2}$ equals the count of carbons.
$H_{2} O$ count $= c = y /2 = 8/2 = 4$ . WHY $c = y /2$ : each water swallows 2 hydrogens, so $y$ hydrogens make $y /2$ waters. Answer: $x = 3, y = 8, z = 0, 3 CO_{2}, 4 H_{2} O$ .

Recall Solution L1.2

Sample A = complete combustion. Every carbon is fully oxidized to $CO_{2}$ and every hydrogen to $H_{2} O$ — the two "full" sinks. Sample B = incomplete combustion. $CO$ (carbon only partly oxidized) and soot (carbon not oxidized at all) appear when there is not enough $O_{2}$ . See Limiting Reagent — oxygen was the limiting reagent.

Level 2 — Application

Now plug numbers into the formula and produce a balanced equation.

Recall Solution L2.1

Use $x = 3, y = 8, z = 0$ .

$b = x = 3$ . WHY: recall $b$ is the $CO_{2}$ coefficient; every carbon becomes one $CO_{2}$ , so $b$ equals the carbon count $x$ .
$c = y /2 = 4$ . WHY: $c$ is the $H_{2} O$ coefficient; two hydrogens make one water, so $c = y /2$ .
$a = x + \frac{y}{4} - \frac{z}{2} = 3 + \frac{8}{4} - 0 = 3 + 2 = 5$ . WHY: $a$ is the $O_{2}$ coefficient, solved last because $O_{2}$ is the only free oxygen source once the products are fixed. $C_{3} H_{8} + 5 O_{2} \to 3 CO_{2} + 4 H_{2} O$ Check oxygen: right side has $2 (3) + 4 = 10$ O atoms; left side $5 \times 2 = 10$ . ✓ (Law of Conservation of Mass holds.)

Recall Solution L2.2

Here $z \neq = 0$ — the fuel carries its own oxygen. $x = 6, y = 12, z = 6$ .

$b = x = 6$ . WHY: $b$ (the $CO_{2}$ coefficient) always equals the carbon count.
$c = y /2 = 12/2 = 6$ . WHY: $c$ (the $H_{2} O$ coefficient) is half the hydrogen count.
$a = 6 + \frac{12}{4} - \frac{6}{2} = 6 + 3 - 3 = 6$ . WHY: $a$ (the $O_{2}$ coefficient) subtracts the fuel's own oxygen last. $C_{6} H_{12} O_{6} + 6 O_{2} \to 6 CO_{2} + 6 H_{2} O$ WHY the $- 3$ ? The $- z /2 = - 6/2 = - 3$ credits the 6 oxygen atoms already sitting inside glucose, so you supply 3 fewer $O_{2}$ than you naively would.

Recall Solution L2.3

$x = 4, y = 10, z = 0$ .

$b = x = 4$ (the $CO_{2}$ coefficient), $c = y /2 = 5$ (the $H_{2} O$ coefficient).
$a = 4 + \frac{10}{4} = 4 + 2.5 = 6.5 = \frac{13}{2}$ (the $O_{2}$ coefficient). WHY the half? With 10 hydrogens the $\frac{y}{4}$ term is not a whole number, so $a$ comes out fractional. $C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 CO_{2} + 5 H_{2} O$ Multiply every coefficient by 2 to clear the half. WHY: balanced equations conventionally use smallest whole numbers, and multiplying all coefficients by the same factor keeps the atom balance intact: $2 C_{4} H_{10} + 13 O_{2} \to 8 CO_{2} + 10 H_{2} O$

Level 3 — Analysis

Now the equation is a tool, and you must convert between moles and grams. See Mole Concept and Molar Mass.

Recall Solution L3.1

Step 1 — moles of fuel. $M (C_{3} H_{8})$ (the molar mass of propane) $= 3 (12) + 8 (1) = 44$ g/mol. So moles $= 22/44 = 0.5$ mol. WHY divide by molar mass? Molar mass is grams-per-mole, so grams ÷ (grams/mole) = moles. Step 2 — mole ratio. From L2.1 the coefficients give $C_{3} H_{8} : O_{2} = 1 : 5$ . So $O_{2} = 0.5 \times 5 = 2.5$ mol. WHY must the ratio be applied in mole-space? Balancing coefficients count molecules (hence moles), not grams — the equation says "1 molecule of propane reacts with 5 molecules of $O_{2}$ ", so we scale moles, never grams, by the ratio. Step 3 — grams. $M (O_{2})$ (the molar mass of oxygen gas) $= 2 \times 16 = 32$ g/mol. Mass $= 2.5 \times 32 = 80$ g. WHY multiply by molar mass? Moles × (grams/mole) = grams — the reverse of Step 1. Answer: $80$ g of $O_{2}$ .

Recall Solution L3.2

Fuel moles: $8.8/44 = 0.2$ mol (grams ÷ molar mass). CO₂: ratio $C_{3} H_{8} : CO_{2} = 1 : 3$ applied in mole-space, so $0.2 \times 3 = 0.6$ mol. $M (CO_{2})$ (molar mass of carbon dioxide) $= 12 + 2 (16) = 44$ g/mol → mass $= 0.6 \times 44 = 26.4$ g. Cross-check with conservation:

$O_{2}$ used $= 0.2 \times 5 = 1.0$ mol $\times 32 = 32$ g.
$H_{2} O$ made $= 0.2 \times 4 = 0.8$ mol $\times 18 = 14.4$ g.
In: $8.8 + 32 = 40.8$ g. Out: $26.4 + 14.4 = 40.8$ g. ✓ Answer: $26.4$ g $CO_{2}$ .

Level 4 — Synthesis

Combine combustion with limiting reagents, unknown formulas, and energy.

Recall Solution L4.1

Moles available: $CH_{4} = 16/16 = 1.0$ mol; $O_{2} = 48/32 = 1.5$ mol (each grams ÷ molar mass). What full reaction of the fuel demands: $1$ mol $CH_{4}$ needs $2$ mol $O_{2}$ . We only have $1.5$ mol $O_{2}$ — oxygen runs out first, so $O_{2}$ is the limiting reagent (Limiting Reagent). Work from the limiter: $1.5$ mol $O_{2}$ × (1 mol $CO_{2}$ / 2 mol $O_{2}$ ) $= 0.75$ mol $CO_{2}$ . WHY work from the limiter? The reaction stops when the scarce reagent is gone, so the product is set by that reagent. Mass $= 0.75 \times 44 = 33$ g. Answer: $O_{2}$ limits; $33$ g $CO_{2}$ form (and $0.25$ mol methane is left unburned — a real-engine cause of incomplete combustion).

Recall Solution L4.2

Carbon: all C is now in $CO_{2}$ . Moles $CO_{2} = 17.6/44 = 0.4$ mol → $0.4$ mol C atoms. WHY: every carbon in the fuel became one $CO_{2}$ , so counting $CO_{2}$ counts the carbons. Hydrogen: all H is now in $H_{2} O$ . Moles $H_{2} O = 9.0/18 = 0.5$ mol → each water holds 2 H, so $0.5 \times 2 = 1.0$ mol H atoms. Ratio C : H $= 0.4 : 1.0 = 2 : 5$ . Empirical formula: $C_{2} H_{5}$ (the real molecule could be $C_{4} H_{10}$ , butane — twice the empirical unit).

Recall Solution L4.3

Moles: $40/16 = 2.5$ mol (grams ÷ molar mass). Heat: energy scales with moles, so $2.5 \times 890 = 2225$ kJ released. WHY multiply? The enthalpy is stated per mole, so total heat = moles × per-mole value. Answer: $2225 kJ$ (i.e. $Δ H = - 2225$ kJ for the batch).

Level 5 — Mastery

Multi-step, mixed-fuel, or "engineer's" problems.

Recall Solution L5.1

O₂ needed: $a = 8 + \frac{18}{4} = 12.5$ mol $O_{2}$ per mole octane (the $O_{2}$ coefficient). Convert to air: if $21%$ of air is $O_{2}$ , then moles of air $= 12.5/0.21 = 59.52$ mol (to 2 dp). WHY divide by 0.21? Only about a fifth of air is oxygen, so you must breathe in roughly five times as much air as the $O_{2}$ you actually need. Answer: $\approx 59.52$ mol of air — this is why engines need such large air intakes.

Recall Solution L5.2

Per-fuel oxygen coefficients (each is the $a$ value for that fuel):

Methane: $a = 1 + 4/4 = 2$ . → $0.3 \times 2 = 0.6$ mol $O_{2}$ ; $CO_{2} = 0.3 \times 1 = 0.3$ mol.
Ethane $C_{2} H_{6}$ : $a = 2 + 6/4 = 3.5$ . → $0.2 \times 3.5 = 0.7$ mol $O_{2}$ ; $CO_{2} = 0.2 \times 2 = 0.4$ mol. Totals: $O_{2} = 0.6 + 0.7 = 1.3$ mol; $CO_{2} = 0.3 + 0.4 = 0.7$ mol. WHY add? Each fuel burns independently, so total oxygen demand and total $CO_{2}$ are the sums of the parts. Answer: $1.3$ mol $O_{2}$ , $0.7$ mol $CO_{2}$ .

Recall Solution L5.3

Carbon: $x = b =$ moles $CO_{2}$ per fuel $= 2$ (the $CO_{2}$ coefficient equals the carbon count). Hydrogen: $c = y /2 = 2 \Rightarrow y = 4$ (the $H_{2} O$ coefficient is half the hydrogen count). Oxygen from molar mass: $12 x + y + 16 z = 60 \Rightarrow 12 (2) + 4 + 16 z = 60 \Rightarrow 28 + 16 z = 60 \Rightarrow 16 z = 32 \Rightarrow z = 2$ . WHY: the molar mass is the sum of each atom's mass; solving it pins down the only unknown $z$ . Formula: $C_{2} H_{4} O_{2}$ — acetic acid (vinegar). Check its $O_{2}$ coefficient: $a = 2 + 4/4 - 2/2 = 2 + 1 - 1 = 2$ mol.

Difficulty ladder at a glance

The next figure captures a genuinely different insight from the ladder: the conversion pipeline every L3–L5 problem quietly runs through — grams enter, become moles, get scaled by a coefficient ratio, and turn back into grams.

Figure — Combustion stoichiometry — fuel + O₂ → CO₂ + H₂O

Recall Master checklist (unfold after finishing all 12)

Balance order is always C, then H, then O last.
Coefficients ( $a, b, c$ ) live in mole-space; convert grams→moles before using them.
The limiting reagent is the smallest (moles ÷ coefficient).
Air needs the extra ÷ 0.21 step for the $O_{2}$ fraction.
Every balanced burn must pass the mass-in = mass-out check.

Connections

1.3.09 Combustion stoichiometry — fuel + O₂ → CO₂ + H₂O (Hinglish) — the parent this drills.
Balancing Chemical Equations — the general skill behind L1–L2.
Mole Concept and Molar Mass — every gram↔mole conversion.
Limiting Reagent — the engine of L4.1.
Enthalpy of Combustion — energy scaling in L4.3.
Law of Conservation of Mass — the cross-checks.
Oxidation States — why $CO_{2}$ is "fully oxidized".