Limiting reagents, yield and purity

Compare what each reactant can support

The limiting reactant is the one exhausted first under the stated complete net-reaction model. Compare the available amounts relative to their required coefficients, not just which mass or mole number is smallest. For 2 H₂ + O₂ → 2 H₂O, a hypothetical mixture has 0.600 mol H₂ and 0.200 mol O₂.

Divide each amount by its coefficient: hydrogen gives 0.600/2 = 0.300 mol of reaction extent, while oxygen gives 0.200/1 = 0.200 mol. The smaller supported extent is 0.200 mol, so O₂ limits this accounting calculation. Equivalently, calculate the possible product from each reactant and select the smaller prediction.

Find product and leftover material

At an extent of 0.200 mol, the equation consumes 0.400 mol H₂ and 0.200 mol O₂, forming 0.400 mol H₂O. Hydrogen remaining is 0.600 − 0.400 = 0.200 mol. The model leaves no oxygen reactant remaining. The table checks all three species together rather than calculating product alone.

This is hypothetical stoichiometry, not a preparation instruction. If measured amount uncertainties are large enough to change which coefficient-adjusted amount is smaller, a unique limiting identity may not be justified. Conversion can also be incomplete. State the idealized assumptions before comparing them with a real experiment.

Yield compares the desired product

Theoretical yield is the maximum desired-product amount or mass predicted from the stated reaction and limiting input. Using M(H₂O) = 18.016 g/mol, 0.400 mol gives 7.2064 g before rounding. If an independently measured recovered mass of the desired pure product is 6.00 g, percent yield is 6.00/7.2064 × 100%, about 83.3%.

A lower recovered yield can reflect incomplete conversion, competing products, separation losses or measurement issues. The yield number alone does not identify the cause. An apparent value over 100% signals that the ideal assumptions or the measured product mass need checking, for example moisture, impurities, incorrect composition or measurement bias. It does not by itself establish extra creation of matter.

Correct the mass before using the equation

A 10.0 g sample stated to contain 80.0% CaCO₃ by mass has 8.00 g of CaCO₃. With M = 100.09 g/mol, its active amount is about 0.0799 mol. For CaCO₃ + 2 HCl → CaCl₂ + CO₂ + H₂O with sufficient acid and inert impurities, that corresponds to the same theoretical CO₂ amount.

Use the mass fraction for the relevant constituent and reaction, not a vague quality label. Impurities may themselves react; if so, the inert-impurity assumption fails. Likewise, a weighed product mixture must be corrected or characterized before its total mass is treated as recovered desired product. Purity, conversion and recovery are related but distinct quantities.

Worked examples

Try it yourself

1. Is the reactant with the smallest mass always limiting?

   Show the answer and reasoning

   No.

   Compare amounts relative to the balanced coefficients and molar masses.

2. For 0.600 mol H₂ and 0.200 mol O₂, what H₂ amount remains under the complete water-reaction model?

   Show the answer and reasoning

   0.200 mol H₂.

   The 0.200 mol O₂ consumes 0.400 mol H₂.

3. What active mass is present in a 10.0 g sample at 80.0% intended substance by mass?

   Show the answer and reasoning

   8.00 g.

   Multiply 10.0 g by mass fraction 0.800 before converting to amount.

4. Does a yield of 83.3% prove a particular side reaction occurred?

   Show the answer and reasoning

   No.

   Yield summarizes the discrepancy; conversion, recovery, composition and measurements need further checks.

What to carry forward

• Compare available amount divided by required coefficient.
• Track reactant remainder as well as product formation.
• Percent yield uses measured desired product relative to theoretical desired product.
• Correct constituent masses for purity within explicit reaction assumptions.