Limiting reagent problems: Worked examples — Chemistry deep dive

Everything here uses two symbols from the parent note, so let us re-anchor them in plain words before we lean on them:

Recall The one number that decides everything

For a reactant $i$ we compute $R_{i} = \frac{n _{i}}{ν _{i}}$ , where $n_{i}$ is how many moles you actually have and $ν_{i}$ is how many moles the balanced recipe eats per reaction-set. So $R_{i}$ = "how many full reaction-sets this reactant can pay for." The smallest $R_{i}$ across all reactants is the limiting reagent. See Limiting reagent problems for the derivation.

The scenario matrix

Every limiting-reagent question you will ever see is one (or a blend) of these cells. Each example below is tagged with the cell it covers.

Cell	Case class	What makes it tricky	Example
A	Standard two-reactant, given in grams	nothing — the baseline	Ex 1
B	Coefficient trap (more moles, still limiting)	a big coefficient eats fast	Ex 2
C	Exact stoichiometric ratio (a "tie")	neither is in excess — a degenerate case	Ex 3
D	Gas given by volume at STP, not grams	must convert litres → moles first	Ex 4
E	Impure reactant (percent purity)	strip out the inert junk before mole-counting	Ex 5
F	Solution reactant (molarity × volume)	moles hide inside "mol/L × L"	Ex 6
G	Real-world word problem (industrial)	translate a paragraph into the 4 steps	Ex 7
H	Exam twist — solve for an unknown mass working backwards from the product	the limiting logic runs in reverse	Ex 8

The three "sign/zero/limit" edge cases in chemistry are:

Zero of a reactant → that reactant is limiting instantly, product = 0 (Ex 3's discussion for a solid, Ex 6's discussion for a solution).
Exact ratio → the tie (Ex 3, Ex 7).
Huge excess (limiting ≈ same as one reactant alone) → Ex 2 & 7.

How to read the figure below

The figure plots the matrix as a bar chart. Along the horizontal axis are five representative cells (A, B, C, D, F). For each cell there are two vertical bars:

the burnt-orange bar = the $R_{i}$ value of the reactant that turns out to be limiting;
the teal bar = the $R_{i}$ value of the other reactant.

The vertical axis is $R = moles / coefficient$ , the exact quantity from the recall box above — so bar height literally means "how many reaction-sets that reactant can fund." The numbers on the bars are the actual $R$ values computed in each example ( $0.50$ for carbon in Cell A, $0.0375$ for the acid in Cell F, and so on). The plum "LIMITS" tag sits over whichever bar is shorter. The single lesson the picture teaches: in every cell the orange (limiting) bar is the shorter one — smallest $R$ wins. Cell C is the exception worth staring at: there the two bars are equal height ( $R = 1.0$ each), the geometric signature of a tie, where neither reactant is in excess.

Example 1 — Cell A: the plain baseline

Forecast: Guess now — carbon has less mass. Does that make it limiting? Write down your guess before reading on.

Balance check. The equation already balances: 1 C, 2 O on each side. Why this step? Coefficients $ν_{i}$ are the "per-set" appetite; a wrong equation poisons every later number.
Grams → moles. $n_{C} = \frac{6.0}{12} = 0.50 mol, n_{O_{2}} = \frac{24.0}{32} = 0.75 mol$ Why this step? Molecules react by count, not weight (see Mole concept and Avogadro number).
Divide by coefficient → $R_{i}$ . $R_{C} = \frac{0.50}{1} = 0.50, R_{O_{2}} = \frac{0.75}{1} = 0.75$ Why this step? $R$ is "reaction-sets funded." $0.50 < 0.75$ , so carbon is limiting.
Scale from the limiter to product. $n_{C O_{2}} = n_{C} \times \frac{1}{1} = 0.50 mol \Rightarrow m = 0.50 \times 44 = 22 g$ Why this step? Only the limiting reagent sets the yield.

Example 2 — Cell B: the coefficient trap

Forecast: Equal moles of each. Feels like a tie, right? Guess who limits.

Both already in moles, so skip the conversion. $n_{F e} = 2.0$ , $n_{O_{2}} = 2.0$ . Why this step? Naive eyes stop here and call it a tie — that is the trap.
Divide by coefficient. $R_{F e} = \frac{2.0}{4} = 0.50, R_{O_{2}} = \frac{2.0}{3} = 0.667$ Why this step? Iron's coefficient is $4$ — the recipe eats iron four at a time, so it drains faster. Fe is limiting despite equal moles.
Product. $n_{F e_{2} O_{3}} = 2.0 \times \frac{2}{4} = 1.0 mol$ Why this step? Scale from limiter using the $F e : F e_{2} O_{3} = 4 : 2$ ratio.

Example 3 — Cell C: the exact-ratio tie (degenerate case)

Forecast: Is there even a leftover here? Guess before computing.

Moles. $n_{H_{2}} = 4.0/2 = 2.0$ , $n_{O_{2}} = 32.0/32 = 1.0$ . Why this step? Standard conversion.
Ratios. $R_{H_{2}} = \frac{2.0}{2} = 1.0, R_{O_{2}} = \frac{1.0}{1} = 1.0$ Why this step? They are equal — this is the degenerate "tie." Neither is in excess; both finish at the same instant. This is the boundary case between " $H_{2}$ limits" and " $O_{2}$ limits."
Product (either reactant gives the same answer — that is what "tie" means). $n_{H_{2} O} = 2.0 \times \frac{2}{2} = 2.0 mol \Rightarrow m = 2.0 \times 18 = 36 g$ Why this step? In a tie either reactant is "limiting," so we scale from $H_{2}$ using the $H_{2} : H_{2} O = 2 : 2$ ratio; scaling from $O_{2}$ (via $O_{2} : H_{2} O = 1 : 2$ ) gives the identical $2.0 mol$ , which is exactly what "tie" guarantees.

Example 4 — Cell D: a gas given by volume (STP)

Forecast: One is a volume, one is a mass. You cannot compare them until both are moles.

Volume → moles (this is the whole point of Cell D). $n_{N_{2}} = \frac{11.2}{22.4} = 0.50 mol$ Why this step? A litre of gas is a count of molecules at fixed T,P (Avogadro). $22.4 L/mol$ is the STP conversion — see Mole concept and Avogadro number.
Mass → moles. $n_{H_{2}} = 5.0/2 = 2.5 mol$ . Why this step? The two reactants arrived in different costumes (one litres, one grams); we must express both in the common currency of moles before any comparison is meaningful.
Ratios. $R_{N_{2}} = \frac{0.50}{1} = 0.50, R_{H_{2}} = \frac{2.5}{3} = 0.833$ Why this step? Dividing each mole count by its coefficient gives "reaction-sets funded"; $H_{2}$ 's coefficient is $3$ , so its supply must be split three ways per set. $0.50 < 0.833$ ⇒ == $N_{2}$ is limiting.==
Product. $n_{N H_{3}} = 0.50 \times \frac{2}{1} = 1.0 mol$ . Why this step? Only the limiting reagent sets the yield; we scale $N_{2}$ up to $N H_{3}$ using the $N_{2} : N H_{3} = 1 : 2$ mole ratio from the balanced equation.

Example 5 — Cell E: impure reactant (percent purity)

Forecast: Do you use $50 g$ or something smaller? Guess.

Strip out the junk first. $m_{C a C O_{3}} = 50.0 \times \frac{80}{100} = 40.0 g$ Why this step? Sand does not react; only the pure portion has a mole-count that matters. Feeding the impure mass in would over-count product (see Percent yield and purity).
Moles of the pure part. $n_{C a C O_{3}} = 40.0/100 = 0.40 mol$ . Why this step? The recipe counts particles, so we convert the reacting mass (the $40.0 g$ , not the $50.0 g$ ) into moles before scaling to product.
Single reactant ⇒ it is "limiting" by default. Scale to product: $n_{C O_{2}} = 0.40 \times \frac{1}{1} = 0.40 mol \Rightarrow m = 0.40 \times 44 = 17.6 g$ Why this step? With only one decomposing reactant there is nothing else to run out, so it caps the yield itself; we scale via the $C a C O_{3} : C O_{2} = 1 : 1$ ratio.

Example 6 — Cell F: reactant given as a solution (molarity)

Forecast: Molarity means "mol per litre." Guess who runs out.

Moles from molarity × volume (Cell F's core move). Convert mL → L first. $n_{N a O H} = 1.0 \times 0.100 = 0.100 mol$ $n_{H_{2} S O_{4}} = 0.50 \times 0.075 = 0.0375 mol$ Why this step? $mol = mol/L \times L$ — see Concentration and molarity.
Ratios. $R_{N a O H} = \frac{0.100}{2} = 0.0500, R_{H_{2} S O_{4}} = \frac{0.0375}{1} = 0.0375$ Why this step? Each reaction-set eats two $N a O H$ per one $H_{2} S O_{4}$ , so we divide the base's moles by $2$ to compare fairly; the smaller quotient, $0.0375$ , marks == $H_{2} S O_{4}$ as limiting== (the acid runs out first).
Product. $n_{N a_{2} S O_{4}} = 0.0375 \times \frac{1}{1} = 0.0375 mol$ . Why this step? Only the limiting acid sets the yield; we scale it to $N a_{2} S O_{4}$ via the $H_{2} S O_{4} : N a_{2} S O_{4} = 1 : 1$ ratio.

Example 7 — Cell G: industrial word problem

Forecast: $1400 kg$ vs $300 kg$ — nitrogen's mass dwarfs hydrogen's. Does that settle it? Guess.

Convert kilograms to kilomoles carefully. Watch the units: molar mass is in $g/mol$ , but our masses are in $kg$ . Since $1 kg = 1000 g$ , $n_{N_{2}} = \frac{1400 kg \times 1000 \frac{g}{kg}}{28 \frac{g}{mol}} = 50, 000 mol = 50 kmol$ $n_{H_{2}} = \frac{300 \times 1000}{2} = 150, 000 mol = 150 kmol$ Why this step? The $1000 \times$ factor (kg→g) applies to both reactants equally, so — as a shortcut — "kg ÷ (g/mol)" gives kilomoles directly. But you must know the $1 0^{3}$ is hiding there, or a mixed-unit problem will bite you.
Ratios. $R_{N_{2}} = \frac{50}{1} = 50, R_{H_{2}} = \frac{150}{3} = 50$ Why this step? Dividing each feedstock's kmol by its coefficient ( $1$ for $N_{2}$ , $3$ for $H_{2}$ ) gives "reaction-sets funded"; the two come out equal, another tie — the plant deliberately feeds the exact $1 : 3$ mole ratio for zero waste, an industrial design choice.
Theoretical product. $n_{N H_{3}} = 50 \times \frac{2}{1} = 100 kmol \Rightarrow m = 100 kmol \times 17 \frac{g}{mol} = 1700 kg$ Why this step? Theoretical yield is scaled from the limiting reagent (either one, since it is a tie) using the $N_{2} : N H_{3} = 1 : 2$ ratio; multiplying kmol by $M_{N H_{3}}$ (with the same kg↔g bookkeeping) returns kilograms.

Example 8 — Cell H: the exam twist (work backwards)

Forecast: Here the product is fixed and we solve for the inputs — the logic runs in reverse. Guess which mass is larger.

From product back to the limiting reactant. Since $H C l$ is chosen as limiting and $H_{2} : H C l = 1 : 2$ : $n_{H C l} = 0.10 \times \frac{2}{1} = 0.20 mol \Rightarrow m_{H C l} = 0.20 \times 36.5 = 7.3 g$ Why this step? The limiting reagent's amount is dictated by the required product — invert the scaling arrow (product → limiter instead of limiter → product).
Zinc needed for exact stoichiometry ( $Z n : H C l = 1 : 2$ ): $n_{Z n, react} = 0.20 \times \frac{1}{2} = 0.10 mol$ Why this step? This is the amount of $Z n$ that would exactly match the $H C l$ — the break-even point where a tie would occur. We want $Z n$ above this so $H C l$ genuinely runs out first.
Add the $20%$ excess to guarantee $H C l$ runs out first. $n_{Z n} = 0.10 \times 1.20 = 0.12 mol \Rightarrow m_{Z n} = 0.12 \times 65 = 7.8 g$ Why this step? " $20%$ excess" means take $120%$ of the exact requirement. This lifts $R_{Z n}$ above $R_{H C l}$ , cementing $H C l$ as the limiting reagent by design.

Recall Scenario self-test

Which cell is " $150 mL$ of $2 M$ HCl reacts with $2 g$ Zn"? ::: Cell F (solution × molarity) blended with Cell A. In a tie, how much of each reactant is left over? ::: Zero of both — they finish simultaneously. What must you do to a $70%$ -pure sample before counting moles? ::: Multiply its mass by $0.70$ to get the reacting mass (Cell E). To convert a gas volume at STP to moles, divide by ::: $22.4 L/mol$ (Cell D). If a reactant is supplied at $0 M$ concentration, how much product forms? ::: Zero — its $R = 0$ , so it is limiting instantly (Cell F edge case). When you convert kg to moles using a molar mass in g/mol, what hidden factor must you include? ::: The $1000 \times$ (kg → g) conversion (Cell G). In the "work backwards" twist, you scale from what? ::: From the fixed target product, back to the limiting reactant (Cell H).

Connections

Limiting reagent problems — the parent method this page drills.
Balancing chemical equations — supplies every $ν_{i}$ used above.
Mole concept and Avogadro number — grams, litres and molarity all funnel into moles.
Stoichiometric calculations — the mole-ratio scaling in every step.
Percent yield and purity — Cell E (impurity) and theoretical-yield logic.
Concentration and molarity — Cell F (solution reactants).
1.3.03 Limiting reagent problems (Hinglish) — same ideas in Hinglish.

Worked examples — Limiting reagent problems