Limiting reagent problems: Question bank — Chemistry deep dive

Before you start, load the one symbol used throughout. The picture below is the whole idea in one glance: each reactant funds a certain number of complete "recipe-runs", and the shortest bar wins.

The second figure shows why the reaction stops: as it runs, both reactants drain, but the limiting one hits zero first and freezes everything — leaving the excess as a leftover pile.

True or false — justify

Statement (commit before revealing)	Verdict + reasoning
Fewer grams always means limiting ::: False. Grams ignore molar mass; a light molecule packs more particles per gram. Convert to moles and divide by the coefficient before comparing.
Fewer moles always means limiting ::: False. You still divide by the coefficient. A reactant with more moles can be limiting if its coefficient is large — the recipe demands many of it per run.
The limiting reagent is always the one with the smallest $R_{i} = n_{i} / ν_{i}$ ::: True. $R_{i}$ counts fundable recipe-runs; the reaction can only run as many times as the poorest funder allows, so the minimum wins.
If two reactants are supplied in exactly the stoichiometric ratio, they run out together ::: True. They hit zero simultaneously, so there is no excess and both cap the yield equally; by convention either one may be labelled the limiting reagent.
Doubling the amount of the excess reagent increases the product ::: False. Product is fixed by the limiting reagent; extra excess just enlarges the leftover pile and does nothing.
The theoretical yield is the largest product you could ever get in the lab ::: True. Computed from the limiting reagent assuming 100% conversion; real yield is at most this, giving percent yield $\leq 100%$ .
Percent yield can exceed 100% for a correctly performed clean reaction ::: False. Above 100% signals error — usually leftover solvent, water, or impure product weighed in with the sample.
Scaling a balanced equation ( $N_{2} + 3 H_{2} \to 2 N H_{3}$ into $2 N_{2} + 6 H_{2} \to 4 N H_{3}$ ) changes the limiting reagent ::: False. Both $n_{i}$ and $ν_{i}$ scale together, so every $R_{i}$ is unchanged — the smallest is still the same reactant.

Spot the error

Flawed claim	Where it breaks
" $50 g$ of A vs $80 g$ of B, so B is in excess." ::: Masses were compared directly. Excess/limiting is decided by $R_{i} = n_{i} / ν_{i}$ ; the heavier mass can still be limiting, as with Al in Worked Example 2.
" $N_{2}$ is limiting, so I'll compute $N H_{3}$ from the leftover $H_{2}$ ." ::: Product must be scaled from the limiting reagent ( $N_{2}$ ), never the excess. Leftover $H_{2}$ cannot react without a partner, so it cannot set the yield.
"I have equal moles of both reactants, so neither is limiting." ::: Equal moles is not equal $R_{i}$ unless the coefficients are equal. Divide each by its own $ν_{i}$ first; the larger-coefficient reactant is likely limiting.
"The equation isn't balanced yet, but I'll compare $R_{i}$ anyway." ::: Coefficients $ν_{i}$ are only meaningful once the equation is balanced. Unbalanced coefficients give a false ratio and the wrong limiting reagent.
"Theoretical yield $34 g$ , actual $34 g$ , so percent yield is undefined." ::: It is simply $\frac{34}{34} \times 100 = 100%$ — a perfectly defined (and maximal) result, not undefined.
"Percent yield is $\frac{theoretical}{actual} \times 100%$ ." ::: The fraction is inverted. Percent yield $= \frac{actual}{theoretical} \times 100%$ ; the actual (smaller) amount goes on top.
"To find leftover excess, subtract limiting-reagent moles from excess-reagent moles." ::: You subtract the moles of excess actually consumed (found via the mole-ratio to the limiting reagent), not the limiting reagent's own moles.

Why questions

Question	Reasoning
Why convert to moles before deciding the limiting reagent, not use grams? ::: Reactions occur in whole-number particle ratios (Mole concept and Avogadro number); grams hide those counts because molar masses differ. Moles are the recipe's true currency.
Why divide moles by the coefficient instead of just comparing moles? ::: The coefficient is how many of that species one recipe-run demands. Dividing gives fundable runs; without it you would penalise a reactant merely for having a large coefficient.
Why does the reaction stop the instant one reactant hits zero? ::: Molecules react only in fixed whole-number groups (Dalton). With no molecules of one species left, no further complete group can form, so the rest just sits.
Why is theoretical yield always computed from the limiting reagent? ::: The limiting reagent runs out first and thereby caps how many recipe-runs occur; product count is set by the number of runs, so only it decides the maximum.
Why can't excess reagent contribute extra product? ::: Every product molecule needs a full set of reactants. Once the limiting partner is gone, excess molecules have nothing to combine with and stay unreacted.
Why does percent yield fall below 100% in real labs even with the right limiting reagent? ::: Side reactions, incomplete conversion, and losses during transfer/filtration mean actual $<$ theoretical, so the ratio drops below 100%.

Edge cases

Scenario	What actually happens
A reactant is supplied at exactly $0 mol$ ::: It is instantly the limiting reagent ( $R_{i} = 0$ ), and zero product forms — no reaction can proceed without it.
Reaction $A + A \to A_{2}$ with only one reactant present ::: No comparison exists; a single reactant simply sets the yield directly. "Limiting reagent" is a concept for two or more reactants in non-stoichiometric ratio.
Two reactants give identical $R_{i}$ values ::: Both run out together; there is no leftover of either. Either may be named limiting, and no reactant is in excess.
A catalyst is present in small amount ::: It is never the limiting reagent. A catalyst is not consumed, so it never runs out and never enters the $R_{i}$ comparison; only true reactants count.
The product can react further (reversible or consecutive reaction) ::: The simple rule gives only the theoretical one-step maximum; equilibrium or further reaction lowers the real yield, captured by percent yield.
A solution reaction — do we compare volumes? ::: No — convert volume and molarity to moles first ( $n = M \times V$ , see Concentration and molarity), then apply $R_{i}$ . Volumes alone, like grams, mislead.

Recall One-line self-test before you leave

If you can answer "why divide moles by the coefficient?" in a single clean sentence, you own this page. Answer ::: Because the coefficient is how many of that species one recipe-run needs, so $n_{i} / ν_{i}$ is the number of runs it can fund — and the fewest runs wins.

Connections

Limiting reagent problems — parent method these traps stress-test.
Balancing chemical equations — where the coefficients $ν_{i}$ come from.
Mole concept and Avogadro number — why moles, not grams, decide.
Stoichiometric calculations — mole-ratio scaling behind every answer.
Percent yield and purity — the yield True/False items feed here.
Concentration and molarity — the solution-phase edge case.