2.2 Stoichiometry, Limiting and Excess Reactants

Once units and stream variables are under control, reactive process calculations add one more layer: stoichiometry. Domain A of the UPDA/MMUP Chemical exam routinely tests whether you can identify the limiting reactant, compute conversion, and interpret yield or selectivity definitions under exam time pressure. The arithmetic is elementary, but the definitions are easy to mix up—especially when feeds are given by mass.

Balancing chemical equations

A balanced reaction conserves every element. For $a\,\mathrm{A} + b\,\mathrm{B} \rightarrow c\,\mathrm{C} + d\,\mathrm{D},$ the stoichiometric coefficients $a,b,c,d$ are pure numbers (often reduced to integers). They define molar ratios only: $\frac{n_{\mathrm{A,\,reacted}}}{a} = \frac{n_{\mathrm{B,\,reacted}}}{b} = \frac{n_{\mathrm{C,\,formed}}}{c} = \xi,$ where $\xi$ is the extent of reaction (moles of reaction events). Extent is the cleanest bookkeeping variable for multi-reaction systems, but single-reaction UPDA items usually work directly with mole tables.

Procedure for any stoichiometry problem

1. Write and balance the reaction.
2. Convert all feed quantities to moles (or molar rates).
3. Divide each reactant’s available moles by its stoichiometric coefficient to find which reactant can support the fewest reaction extents—that reactant is limiting.
4. Compute theoretical product for complete conversion of the limiting reactant.
5. Apply the stated conversion, yield, or selectivity to get actual product and residual reactants.

Limiting reactant, excess reactant, and percent excess

Suppose feed supplies $n_{\mathrm{A0}}$ and $n_{\mathrm{B0}}$ moles of A and B for the reaction $a\mathrm{A} + b\mathrm{B} \rightarrow$ products. The stoichiometric requirement of B for complete use of A is $n_{\mathrm{B,\,req}} = n_{\mathrm{A0}} \times \frac{b}{a}.$

• If $n_{\mathrm{B0}} < n_{\mathrm{B,,req}}$, B is limiting and A is in excess.
• If $n_{\mathrm{B0}} > n_{\mathrm{B,,req}}$, A is limiting and B is in excess.

Percent excess of B (when A is limiting) is $\%\,\mathrm{excess\,B} = \frac{n_{\mathrm{B0}} - n_{\mathrm{B,\,req}}}{n_{\mathrm{B,\,req}}} \times 100\%.$ Some exam writers give “20% excess air” for combustion; that means air moles fed are 1.20 times the stoichiometric air calculated for the fuel that actually reacts (or for complete fuel combustion—read carefully).

Yield and selectivity wording varies. Before computing, underline whether the denominator is “A fed,” “A reacted,” or “theoretical maximum P.” UPDA-style MCQs often hinge on that single phrase.

Fractional conversion

For reactant A: $X_A = \frac{\text{moles of A reacted}}{\text{moles of A fed}} = \frac{n_{\mathrm{A0}} - n_A}{n_{\mathrm{A0}}}.$ Conversion is defined for a reactant, not for a product. Saying “50% conversion of product” is meaningless. In a continuous reactor, use molar flow rates $\dot{n}$ with the same formula. Incomplete conversion leaves unreacted limiting reactant in the outlet even if another species was fed in excess.

Worked example 1 — Limiting reactant and percent excess

Ammonia is produced by $\mathrm{N_2} + 3\mathrm{H_2} \rightarrow 2\mathrm{NH_3}$. A feed contains 100 kmol/h N₂ and 250 kmol/h H₂. Identify the limiting reactant, the percent excess of the other reactant, and the maximum NH₃ production rate if conversion of the limiting reactant is 100%.

Solution. Required H₂ for 100 kmol/h N₂: $100 \times 3 = 300,\mathrm{kmol/h}$. Only 250 kmol/h H₂ is available, so H₂ is limiting and N₂ is in excess.

N₂ required by the available H₂: $250 \times (1/3) = 83.33,\mathrm{kmol/h}$. $\%\,\mathrm{excess\,N_2} = \frac{100 - 83.33}{83.33} \times 100\% = 20\%.$ Maximum NH₃ at complete H₂ conversion: $250 \times (2/3) = 166.7,\mathrm{kmol/h}$.

If someone wrongly treated N₂ as limiting because “less N₂ is fed,” they would report 200 kmol/h NH₃ and miss the answer. Always compare moles-fed-per-coefficient: N₂ → $100/1 = 100$; H₂ → $250/3 = 83.3$; smaller number wins (H₂).

Worked example 2 — Conversion with excess air (combustion style)

Carbon burns: $\mathrm{C} + \mathrm{O_2} \rightarrow \mathrm{CO_2}$. Feed: 50 kmol C/h with 20% excess air. Air is 21 mol% O₂ and 79 mol% N₂. Assume complete conversion of carbon. Find O₂ fed, N₂ fed, and dry product composition.

Solution. Stoichiometric O₂ = 50 kmol/h. With 20% excess: $\mathrm{O_2\,fed} = 1.20 \times 50 = 60\,\mathrm{kmol/h}.$ Air fed = $60 / 0.21 = 285.7,\mathrm{kmol/h}$; N₂ fed = $0.79 \times 285.7 = 225.7,\mathrm{kmol/h}$.

After complete C combustion: CO₂ = 50, O₂ remaining = $60 - 50 = 10$, N₂ = 225.7 (inert). Dry product total = $50 + 10 + 225.7 = 285.7,\mathrm{kmol/h}$. Mole fractions (dry): CO₂ 0.175, O₂ 0.035, N₂ 0.790.

Worked example 3 — Yield and selectivity

Reactant A can form desired P or undesired U: $\mathrm{A} \rightarrow \mathrm{P}, \qquad \mathrm{A} \rightarrow \mathrm{U}.$ Feed 100 mol A; outlet: 40 mol A, 45 mol P, 15 mol U. Find conversion of A, yield of P based on A fed, and selectivity of P relative to U.

Solution. A reacted = $100 - 40 = 60$ mol → $X_A = 0.60$. Yield of P on A fed = $45/100 = 0.45$ (45%). Selectivity $S_{P/U} = 45/15 = 3.0$ (moles P per mole U). Note yield based on A reacted would be $45/60 = 0.75$—a different number. MCQ stems that omit the basis of yield are incomplete; when both definitions appear as options, read the stem’s exact wording.

Worked example 4 — Mass trap

The reaction $2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}$ is run with 4 kg H₂ and 32 kg O₂. Who is limiting?

Solution in moles: H₂ = $4/2 = 2,\mathrm{kmol}$; O₂ = $32/32 = 1,\mathrm{kmol}$. Stoichiometric need: 2 kmol H₂ requires 1 kmol O₂—exact stoichiometric feed, neither excess. If you compared 4 kg vs 32 kg by mass alone, you would falsely call H₂ limiting. Always convert to moles first.

Exam traps to memorize

1. Mass used as moles — especially with light gases (H₂, He) or heavy organics.
2. Wrong limiting reactant — forgetting to divide by stoichiometric coefficients.
3. % excess based on wrong reactant — excess is relative to the stoichiometric partner of the limiting species.
4. Conversion applied to excess reactant — conversion statements usually refer to the limiting reactant or a named species; apply $X$ only to that species.
5. Yield vs selectivity confusion — yield compares product to reactant; selectivity compares product pathways.
6. Unbalanced equation — one missing atom invalidates every ratio downstream.

Master these definitions now; Chapter 3 will embed them inside multi-unit material balances with recycle and purge, where a single stoichiometry error propagates through the entire flowsheet.