Product Amount from the Limiting Reagent

Introduction

Finding a limiting reagent is a decision point, not the final answer. The product calculation must use the reagent that limits reaction progress. A larger prediction from the other reagent describes what could happen only if more of the limiting reagent were supplied.

Core explanation

Take N₂ + 3H₂ → 2NH₃ with 0.50 mol N₂ and 0.90 mol H₂. Nitrogen alone could support 0.50 × 2/1 = 1.00 mol NH₃. Hydrogen alone could support 0.90 × 2/3 = 0.60 mol NH₃. Both cannot be correct for the same starting mixture. Hydrogen supplies the smaller product capacity and is therefore limiting, so 0.60 mol NH₃ is the theoretical maximum. Calculating from nitrogen after identifying hydrogen as limiting would ignore the shortage and overstate production.

The amount-over-coefficient method reaches the same result. n(N₂)/1 = 0.50 mol of reaction, while n(H₂)/3 = 0.30 mol of reaction. The maximum extent is 0.30 mol. Multiplying by the NH₃ coefficient 2 gives 0.60 mol product. Nitrogen consumed is 0.30 mol, leaving 0.20 mol in the ideal account. The two methods are equivalent: comparing possible product amounts is comparing reaction capacities multiplied by the same product coefficient.

If mass is requested, convert the theoretical product moles using the product's molar mass. With M(NH₃) ≈ 17.03 g mol⁻¹, 0.60 mol NH₃ corresponds to about 10.2 g theoretical NH₃. That number does not mean the experiment necessarily recovers 10.2 g; actual yield may be lower. The limiting-reagent concept supplies an upper bound under the chosen reaction and supplies, while reaction equilibrium, losses and side reactions affect actual recovery.

Some equations have multiple products. In CaCO₃ → CaO + CO₂, there is only one reactant, so its available amount limits both products in the 1:1:1 ratio. In a reaction with two reactants and two products, identify the limiting reactant once, then use its coefficient ratio separately for each named product. For example, 2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2CO₂ + 2H₂O gives two moles CO₂ and one mole Na₂SO₄ per two moles bicarbonate. The same limiting extent produces different amounts of the two products.

The basis must be chemically correct. A measured 5.0 g of impure reagent is not 5.0 g of the reacting compound; correct for purity first. A solution concentration refers to the named solute, not the entire mixture. If gaseous reagents are compared by volume, ensure matching conditions or convert each through its own P and T. Only then determine the limiting supply and calculate product.

Close quantities warrant attention to measurement precision. If two reactant capacities are equal within the uncertainty of the supplied measurements, the problem may intend a stoichiometric mixture, or the limiting identity may be ambiguous experimentally. Carry guard digits and compare unrounded values. A clear theoretical calculation should state the equation, the limiting reagent and the assumption of complete reaction of that reagent.

Step-by-step reasoning

1. Balance the equation and convert every given reactant supply into moles. 2. Compare n/coefficient for all reactants or calculate each possible product capacity. 3. Choose the smallest capacity and name its limiting reactant. 4. Multiply its amount by the product-to-limiting-reactant coefficient ratio. 5. Convert product moles to the requested unit and label the result theoretical.

Visual explanation

Show two arrows into one product box. An arrow from 0.50 mol N₂ proposes 1.00 mol NH₃; an arrow from 0.90 mol H₂ proposes 0.60 mol NH₃. Put a stop symbol at the larger proposal and mark the smaller as the supply-constrained result. This makes the comparison meaningful rather than merely procedural.

Real-world analogy

Two resources may each suggest a different maximum number of finished kits. If one supports six kits and the other ten, only six can be completed. The ten-kit estimate assumes additional units of the scarce resource. A limiting-reagent product calculation uses the smaller feasible production count.

Real-world example

For 2Mg + O₂ → 2MgO, 0.30 mol Mg and 0.10 mol O₂ are supplied. Magnesium could make 0.30 mol MgO, while oxygen could make only 0.20 mol MgO. Oxygen limits; the theoretical oxide amount is 0.20 mol, or about 8.06 g using M(MgO) ≈ 40.31 g mol⁻¹.

Why?

Why discard the larger product prediction? It is conditional on consuming all of a reagent that actually remains in excess. The insufficient co-reactant prevents that full consumption, so only the smaller capacity is feasible for the supplied mixture and balanced equation.

Common misconception

“Add the product amounts predicted independently from each reactant.” Those predictions refer to the same reaction, not separate batches. Adding them double-counts the available materials and violates the requirement that each product unit consume the full balanced set of reactants.

Worked example

Find the theoretical mass of Fe₂O₃ from 0.50 mol Fe and 0.30 mol O₂. Balance 4Fe + 3O₂ → 2Fe₂O₃. Iron capacity is 0.50/4 = 0.125 mol of reaction; oxygen capacity is 0.30/3 = 0.100 mol. Oxygen limits. Product amount is 2 × 0.100 = 0.200 mol Fe₂O₃. With M(Fe₂O₃) = 159.70 g mol⁻¹, theoretical mass is 31.9 g to three significant figures. Iron consumed is 4 × 0.100 = 0.400 mol, leaving 0.100 mol Fe in the ideal account. Calculating from all 0.50 mol Fe would suggest 0.250 mol oxide, but that would require 0.375 mol O₂, more than the 0.30 mol supplied.

Quick check

1. What theoretical Fe₂O₃ amount follows when oxygen limits the reaction extent to 0.100 mol as written? Answer: The product coefficient is two, so the maximum is 0.200 mol Fe₂O₃ formula units.

Exam focus

Show both reactant capacities before using a product factor. The numerator of the final ratio must be the named product and the denominator the identified limiting reagent. State whether the answer is moles or grams and call it theoretical unless actual recovery data are supplied.

Advanced insight

For product P with coefficient νP and maximum extent ξmax, nP,theoretical = νPξmax. This compact formula handles any number of products once ξmax has been determined by the smallest reactant nᵢ/ νᵢ . It also separates production bounds from questions of reaction kinetics or equilibrium.

Summary

The limiting reagent alone sets the maximum product from a supplied mixture. Compare every reactant's reaction capacity, use the smallest, then multiply by the product coefficient and any required molar-mass factor. Larger predictions from excess reagents are counterfactual for the actual starting supplies.

Practice questions

1. For N₂ + 3H₂ → 2NH₃, what maximum NH₃ forms from 0.50 mol N₂ and 0.90 mol H₂? Answer: 0.60 mol NH₃; hydrogen is limiting. 2. In 2Mg + O₂ → 2MgO, which limits 0.30 mol Mg and 0.10 mol O₂? Answer: Oxygen limits because it supports only 0.20 mol MgO. 3. Why not add product capacities calculated from two reactants? Answer: Both capacities describe the same batch and would double-count its potential production. 4. What theoretical oxide amount follows from ξmax = 0.10 mol when product coefficient is two? Answer: 0.20 mol oxide formula units. 5. What mass is 0.200 mol Fe₂O₃ if M = 159.70 g mol⁻¹? Answer: 31.9 g to three significant figures.