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General Chemistry · Stoichiometry

Limiting Reagents

Set the masses of two reactants, predict which one runs out first, then reveal the limiting reagent, the mass of product formed, and how much of the excess reactant is left over. Finish by combining limiting-reagent analysis with percent yield.

Theory — Limiting and Excess Reagents

Why One Reactant Runs Out First

When two reactants are mixed, they are rarely present in the exact ratio the balanced equation requires. The limiting reagent is the reactant that is used up first — it limits how much product can form. The other reactant is the excess reagent; some of it is left over when the reaction stops.

Finding the Limiting Reagent 1. Balance the equation.
2. Convert each reactant mass to moles (mass ÷ molar mass).
3. Divide each by its coefficient — the smallest value is the limiting reagent.
4. Use the limiting reagent's moles to find the product.
The limiting reagent determines the theoretical yield

Product Formed and Excess Remaining

The amount of product is calculated from the limiting reagent only. The excess remaining is found by subtracting the amount of excess reactant actually used from the amount supplied.

Key Calculations moles of product = moles of limiting reagent × (product coeff ÷ limiting coeff)
mass of product = moles of product × molar mass
excess used = moles of limiting × (excess coeff ÷ limiting coeff) × molar mass
excess remaining = excess supplied − excess used
Percent yield = (actual ÷ theoretical) × 100

Limiting Reagent

Runs out first; sets the maximum product. Found by the smallest moles ÷ coefficient.

Excess Reagent

Left over after the reaction. Excess remaining = supplied − used.

Percent Yield

Compares the actual product obtained to the theoretical maximum.

StepWhat you do
1. BalanceWrite the balanced equation
2. Molesmass ÷ molar mass for each reactant
3. Comparemoles ÷ coefficient — smallest is limiting
4. Productfrom limiting reagent moles
5. Excesssupplied − used

Apparatus

The equipment a real limiting-reagent experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.

0.000 gmeasures mass
Analytical balance
Weighs each reactant to set the starting amounts.
holds solutions
Beaker
Where the two reactants combine.
reagent solutions
Reagent bottles
Hold the reactant solutions added in measured amounts.
measures volume
Graduated cylinder
Measures reactant solution volumes.
filters solids
Filter funnel
Collects the product to measure the actual yield.
reaction flask
Reaction flask
Holds the mixture as the product forms.

Instructions — Running the Virtual Experiment

The Limiting Reagent bench lets you set two reactant masses and work out which is limiting; the Percent Yield section adds an actual yield. Record every reading in your lab report with screenshots.

Part 1 — Find the Limiting Reagent (Limiting Reagent tab)
1
Open Simulation → Limiting Reagent. Choose a reaction and set the mass of each reactant. Predict which reactant is limiting using the two prediction buttons.
2
Click Run reaction to reveal the limiting reagent, the mass of product formed, and the mass of excess reactant remaining. Record all three for at least three different mass combinations.
Part 2 — Percent Yield (Percent Yield tab)
1
Open Percent Yield. Choose a reaction. You are given the reactant amounts and the experimental (actual) yield. Work out the theoretical yield from the limiting reagent, then calculate the percent yield yourself.
2
Enter your calculated percent yield and click Check my answer to confirm it (use Show solution if needed). Record the limiting reagent, theoretical yield, experimental yield, and percent yield for at least two reactions.

Simulation — Limiting Reagents

Limiting Reagents Virtual LabPredict the limiting reagent, then run the reaction
2 Mg + O₂ → 2 MgO
Mg
4.86 g
O₂
5.00 g
MgO
— g

Predict: which reactant is the limiting reagent?

Reaction & masses

Result
Moles A0.200 mol
Moles B0.156 mol
Limiting reagent
Product formed— g
Excess remaining— g
Set masses, predict, then run the reaction.
2 Mg + O₂ → 2 MgO

You are given the reactant amounts and the experimental (actual) yield obtained in the lab. Work out the theoretical yield from the limiting reagent, then calculate the percent yield and enter it below to check your answer.

Reaction

Given data
Mg supplied4.86 g
O₂ supplied5.00 g
Experimental yield7.10 g
Use these values to calculate the percent yield.

Your calculation

Team Questions

Question 1. In 2Mg + O₂ → 2MgO, you have 0.200 mol Mg and 0.156 mol O₂. Dividing each by its coefficient (Mg ÷ 2, O₂ ÷ 1), which is the limiting reagent? (Type Mg or O₂)
Question 2. Define the excess reagent in one short phrase.
Question 3. For Fe + S → FeS, 5.585 g Fe (0.100 mol) reacts with 6.41 g S (0.200 mol). Which is limiting? (Type Fe or S)
Question 4. In Q3, how many grams of FeS form? Iron is limiting at 0.100 mol; FeS molar mass = 87.92 g/mol. (Type in g, e.g. 8.79)
Question 5. In Q3, how much sulfur (excess) remains? 0.100 mol S is used (1:1), 0.200 mol supplied; S molar mass = 32.07 g/mol. (Type in g, e.g. 3.21)
Question 6. A reaction has a theoretical yield of 8.06 g and an actual yield of 7.00 g. What is the percent yield? (Type to 1 decimal, e.g. 86.8)
Question 7 — Challenge. Does adding more of the excess reagent increase the amount of product formed? (yes or no, with the reason)

Example Lab Report

Sample report demonstrating the expected format. Include labelled screenshots of the bench for each combination, and your full working for the limiting reagent and percent yield.

Limiting Reagents

Chemistry | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To identify the limiting and excess reagents in a reaction from the masses of the reactants, to calculate the mass of product formed and the mass of excess reagent remaining, and to determine the percent yield.

Theory

The limiting reagent is the reactant that is consumed first and sets the theoretical yield. It is found by converting each reactant mass to moles, dividing by its coefficient, and choosing the smallest. The product is calculated from the limiting reagent; the excess remaining is the supplied amount minus the amount used.

moles = mass / molar mass · compare moles ÷ coefficient · product from limiting reagent

Results (worked example: 2Mg + O₂ → 2MgO)

QuantityValue
Mass of Mg4.86 g (0.200 mol)
Mass of O₂5.00 g (0.156 mol)
Mg ÷ 20.100
O₂ ÷ 10.156
Limiting reagentMg
MgO formed (0.200 mol × 40.30)8.06 g
O₂ used (0.100 mol × 32.00)3.20 g
O₂ remaining (5.00 − 3.20)1.80 g

Percent yield example: if the actual MgO obtained is 7.00 g, percent yield = 7.00 / 8.06 × 100 = 86.8 %.

Analysis

Magnesium was limiting because its moles ÷ coefficient (0.100) was smaller than that of oxygen (0.156). All of the magnesium reacted, while 1.80 g of oxygen was left over. Increasing the oxygen further would not change the amount of MgO formed, because magnesium still limits the reaction.

Conclusion

The limiting reagent (magnesium) determined the theoretical yield of magnesium oxide, the excess oxygen remained unreacted, and the percent yield compared the actual product to that theoretical maximum. Identifying limiting and excess reagents is essential for predicting product amounts and using reactants efficiently in real processes.

Practice Questions

Show all work. Find the limiting reagent first, then the product and any excess. Molar masses: Mg 24.31, O₂ 32.00, MgO 40.30, Al 26.98, Cl₂ 70.90, AlCl₃ 133.33, Zn 65.38, HCl 36.46, ZnCl₂ 136.29 g/mol.

Question 1
For 2Mg + O₂ → 2MgO, 6.00 g Mg reacts with 2.00 g O₂. Identify the limiting reagent and find the mass of MgO formed.
Hint: mol Mg = 6.00/24.31; mol O₂ = 2.00/32.00; compare ÷ coefficient.
Question 2
For 2Al + 3Cl₂ → 2AlCl₃, 5.40 g Al reacts with 15.0 g Cl₂. Which is limiting, and how much AlCl₃ forms?
Hint: mol Al/2 vs mol Cl₂/3.
Question 3
In Question 2, how many grams of the excess reagent remain after the reaction?
Hint: excess remaining = supplied − used by the limiting reagent.
Question 4
For Zn + 2HCl → ZnCl₂ + H₂, 6.54 g Zn reacts with 14.58 g HCl. Identify the limiting reagent and the mass of ZnCl₂ formed.
Hint: mol Zn/1 vs mol HCl/2.
Question 5
A reaction has a theoretical yield of 26.7 g of AlCl₃ but only 22.0 g is obtained. What is the percent yield?
Hint: percent yield = actual / theoretical × 100.
Question 6 — Challenge
Explain why, once the limiting reagent is used up, adding more of the excess reagent produces no additional product.
Hint: there is no more limiting reagent left to react with it.