Limiting Reagent & Stoichiometry Lab

Compare moles divided by the balanced coefficient, then sweep one reagent to see yield rise to a kink and flatten.

Worked example — Limiting Reagent & Stoichiometry Lab

reaction batches	0.9921
2NH₃	33.792 g
Smallest moles ÷ coefficient	0.9921
Theoretical yield	33.792 g
Excess left	0.208 g

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

The limiting reagent is not the one you have least of 🖖

With 28 g of N₂ and 6 g of H₂, hydrogen supplies 2.9762 mol against nitrogen's 0.9995 mol, yet H₂ limits. The equation requires three H₂ for every N₂, so the deciding values are 0.9921 versus 0.9995 after dividing by the coefficients.

Yield is two straight lines with a kink 🖖

Sweep either reactant in the chart: yield rises linearly until the two mole-to-coefficient ratios match, then becomes flat. That kink is the stoichiometric point; beyond it, added reagent becomes leftover rather than product.

Above 100% means contamination or measurement error 🖖

The theoretical yield is an upper bound fixed by conservation of atoms. A reported percent yield above 100% therefore does not beat stoichiometry: the collected product is commonly wet or impure, or one of the masses was measured incorrectly.

Problems solved in full

Eight grams of hydrogen, forty of oxygen, and the oxygen runs out first 6 steps
Burn 8 g of hydrogen with 40 g of oxygen: 2H₂ + O₂ → 2H₂O. Find which reagent limits the reaction, the mass of water, and what is left in the vessel. Then work out exactly how much more oxygen would be needed for nothing to be left over.
1. Mass tells you nothing until it is moles, because the equation counts molecules. H₂ is 2 × 1.008 = 2.016 g/mol and O₂ is 2 × 15.999 = 31.998 g/mol.
  
  $M (H_{2}) = 2.016, M (O_{2}) = 31.998 g mo l^{- 1}$
2. Divide. 8 ÷ 2.016 = 3.9683 mol of H₂, and 40 ÷ 31.998 = 1.2501 mol of O₂. There is 3.17 times as much hydrogen by count, from a fifth of the mass.
  
  $n (H_{2}) = \frac{8}{2.016} = 3.9683, n (O_{2}) = \frac{40}{31.998} = 1.2501 mol$
3. Now divide each by its coefficient, which is the step that decides everything: H₂ gives 3.9683 ÷ 2 = 1.9841, and O₂ gives 1.2501 ÷ 1 = 1.2501. The smaller number is the one that runs out, so oxygen is limiting — five times the mass of the other reagent, and it is the one in short supply.
  
  $\frac{3.9683}{2} = 1.9841 > \frac{1.2501}{1} = 1.2501 \Rightarrow O_{2} limits$
4. That smallest ratio, 1.2501, is how many times the whole equation can run. Water comes out at 2 per run, and H₂O is 18.015 g/mol, so the yield is 1.2501 × 2 × 18.015 = 45.04 g.
  
  $m (H_{2} O) = 1.2501 \times 2 \times 18.015 = 45.04 g$
5. Check the vessel. The reaction consumes 2 × 1.2501 = 2.5002 mol of H₂, which is 5.04 g, leaving 2.96 g of hydrogen unburnt. Mass in was 48 g; mass out is 45.04 + 2.96 = 48.00 g, which is the arithmetic checking itself.
  
  $8 - 2 \times 1.2501 \times 2.016 = 2.96 g H_{2} left, 45.04 + 2.96 = 48.00 g$
6. For nothing to be left over, oxygen must reach the hydrogen's ratio of 1.9841, so 1.9841 × 31.998 = 63.49 g. You are 23.49 g short, and adding exactly that much takes the yield to 71.49 g. One gram more than that buys nothing at all.
  
  $1.9841 \times 31.998 = 63.49 g \Rightarrow 23.49 g more, yield 71.49 g$
Answer

Oxygen limits it, the yield is 45.04 g, 2.96 g of hydrogen survives, and 23.49 g more oxygen would be needed to consume it. The trap here is not the arithmetic, it is that mass looks like a quantity of reagent and is not. The vessel holds five times as much oxygen as hydrogen by weight and under a third as much by count, and the equation only counts. Dividing by the coefficient is the step that gets skipped, and skipping it is how you get the answer exactly backwards on a reaction where the numbers are as unbalanced as these. Watch the sweep and the reason for the kink is now visible: below 63.49 g every extra gram of oxygen is another 1.126 g of water, and above it the hydrogen has already gone and the line is flat.
28.4 grams of ammonia collected, an 84% yield, and where the missing 5.4 went 5 steps
Load the preset that carries an actual yield: 28 g of nitrogen, 6 g of hydrogen, 28.4 g of ammonia collected. Work out the percent yield, then decide whether the shortfall could still be sitting in the vessel.
1. The ceiling belongs to whichever reagent runs out first. Six grams of hydrogen is 2.9762 mol, which is 0.9921 runs of the equation once it is divided by the coefficient 3; nitrogen offers 0.9995. Hydrogen limits, by about three quarters of a percent, and 0.9921 is what the smallest moles ÷ coefficient card prints.
  
  $\frac{6}{2.016} = 2.9762 \Rightarrow \frac{2.9762}{3} = 0.9921 < \frac{28}{28.014} = 0.9995$
2. Each run makes two ammonia and NH₃ is 17.031 g/mol, so the most this vessel can produce is 33.792 g. That is a ceiling fixed by the atoms charged. Nothing in it knows whether the reaction goes at all.
  
  $m_{m a x} (N H_{3}) = 0.9921 \times 2 \times 17.031 = 33.792 g$
3. Divide the collected mass by the ceiling and the percent yield is 84.0%. There are 5.392 g unaccounted for.
  
  $η = \frac{28.4}{33.792} = 84.0%, 33.792 - 28.4 = 5.392 g$
4. Could that still be reagent in the vessel? Take the tool at its word: 0.208 g of nitrogen is left over, and even if all of it had reacted it would have added 0.253 g of ammonia, under a twentieth of the gap, with no hydrogen left to react with anyway. But that 0.208 g is itself computed on the assumption that the reaction ran to completion. Since 5.392 g cannot be accounted for under that assumption, the assumption is what failed.
  
  $(0.9995 - 0.9921) \times 28.014 = 0.208 g N_{2} \Rightarrow \frac{0.208}{28.014} \times 2 \times 17.031 = 0.253 g$
5. The ceiling is computed from what was charged, so it moves when the charge does. Put in 6.045 g of hydrogen instead of 6.000, the exact amount 28 g of nitrogen asks for, and nitrogen becomes the limiting reagent, the ceiling rises to 34.045 g, and the same 28.4 g of collected ammonia now reports 83.4%.
  
  $0.9995 \times 3 \times 2.016 = 6.045 g H_{2} \Rightarrow m_{m a x} = 34.045 g \Rightarrow η = \frac{28.4}{34.045} = 83.4%$
Answer

84.0%, and no: at most 0.253 g of the missing 5.392 g had anywhere in the flask to be. What the shortfall shows is that the reaction did not run to completion. The theoretical yield is an atom count and says nothing about how far a reversible reaction travels in one pass, and ammonia synthesis is the standard example of one that does not travel far, which is why industrial plants recycle the unconverted gas instead of accepting the single-pass figure.

The denominator deserves the same suspicion as the numerator, because only the numerator was ever weighed. The bottom of the fraction is inferred from what went in, so two runs collecting an identical 28.4 g report 84.0% or 83.4% depending on how much hydrogen was charged beside the nitrogen. A percent yield quoted without the amounts charged is half a measurement, and it is the half that is easier to move.

$η = \frac{m _{obs}}{ν _{p} M _{p} min _{i} ( n _{i} / ν _{i} )}$

By the 42Math team · Updated 2026-09-03

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Bio, Geo, Chem lessons →

Example problems

Haber knife-edge - H₂ limits at 0.9921 mol per coefficient; NH₃ theoretical yield is 33.792 g and 0.208 g of N₂ remains.
Methane in excess air - CH₄ limits; total product yield is 79.824 g and 64.176 g of O₂ remains.
Mass misleads - The heavier O₂ portion limits; product yield is 45.040 g and 2.960 g of H₂ remains.
With actual yield - An actual yield of 28.4 g against 33.792 g theoretical gives 84.0% yield.