Grade by Grade
D. Quantities in Chemical Reactions · Quantities in Chemical Reactions

Limiting and Excess Reagents

D2.6 D2.1 D3.4 D2.5About 75 minutes
Learning goals
  • I can explain what limiting and excess reagents are.
  • I can identify the limiting reagent from moles or masses of reactants.
  • I can calculate the maximum amount of product from the limiting reagent.
  • I can calculate how much excess reagent is left over.
  • I can explain why industries often add one reactant in excess.

Words to know

limiting reagent (limiting reactant)
The reactant that is used up first. It stops the reaction and decides the maximum amount of product.
excess reagent
A reactant that is present in more than the amount needed; some is left over when the reaction stops.
stoichiometric amounts
Reactant amounts in exactly the mole ratio of the balanced equation, so nothing is left over.
theoretical yield
The maximum amount of product that the limiting reagent can make (more on this next lesson).

Ten slices of bread, three slices of cheese

You are making cheese sandwiches for a picnic. Each sandwich needs 2 slices of bread and 1 slice of cheese. You have 10 slices of bread but only 3 slices of cheese. How many sandwiches can you make? Three, and 4 slices of bread are left over. The cheese limits you: it runs out first, while you still have more bread than you need.

Real reactions are the same. Reactants are rarely mixed in exactly the right ratio. One runs out first and the reaction stops.

Remember from Grade 10

In Grade 10 you met the term limiting reactant in the single displacement lesson: the reactant that runs out first. Now you will identify it with calculations and use it to predict how much product forms. ("Reagent" and "reactant" mean the same thing here.)

10 slices of bread3 slices of cheese3 sandwiches4 bread leftrecipe: 2 bread + 1 cheese → 1 sandwichCheese runs out first: it is the LIMITING reagent. Bread is in EXCESS.
Cheese limits the number of sandwiches. Having more bread does not help once the cheese is gone.

Limiting and excess reagents

  • The limiting reagent is completely used up. It decides the maximum amount of product.
  • The excess reagent is left over when the reaction stops.

Watch out: the limiting reagent is not simply the one with the smaller mass, or even the smaller number of moles. You must compare amounts using the mole ratio from the balanced equation. In the sandwich example, there were fewer slices of cheese than of bread, but that was not the reason it was limiting: the recipe needs 2 bread for every 1 cheese, and 10 bread would need 5 cheese.

before: 6 H₂ + 4 O₂left overafter: 6 H₂O + 1 O₂ (excess)2H₂ + O₂ → 2H₂O: the 6 H₂ need only 3 O₂, so H₂ is limiting.= H atom= O atom
Particles before and after. The equation needs 1 O₂ for every 2 H₂, so 6 H₂ react with 3 of the 4 O₂. Hydrogen runs out; one O₂ molecule is left over.
1. Convert eachreactant to moles2. For ONE reactant, findmoles of the otherit needs (mole ratio)3. Compare neededwith available:who runs out?4. Use the LIMITING reagent's molesto calculate products and leftover excess
A reliable method for every limiting-reagent problem.
Worked example

Example 1: Starting from moles

2H₂(g) + O₂(g) → 2H₂O(g). A mixture contains 5.0 mol of H₂ and 3.0 mol of O₂. Which is the limiting reagent? What amount of water forms? What amount of excess reagent is left?

  1. O₂ needed for all the H₂: 5.0 mol H₂ × (1 mol O₂ ÷ 2 mol H₂) = 2.5 mol O₂.
  2. We have 3.0 mol O₂, which is more than 2.5 mol, so O₂ is in excess and H₂ is limiting.
  3. Water from the limiting reagent: 5.0 mol H₂ × (2 ÷ 2) = 5.0 mol H₂O.
  4. O₂ left over: 3.0 − 2.5 = 0.5 mol.

H₂ is limiting; 5.0 mol of H₂O forms; 0.5 mol of O₂ is left over.

Worked example

Example 2: Starting from masses

Aluminium reacts with chlorine: 2Al(s) + 3Cl₂(g) → 2AlCl₃(s). 10.0 g of Al and 35.0 g of Cl₂ are mixed. Find the limiting reagent, the mass of AlCl₃ formed and the mass of excess reagent left.

  1. Moles: n(Al) = 10.0 ÷ 26.98 = 0.37064 mol. n(Cl₂) = 35.0 ÷ 70.90 = 0.49365 mol.
  2. Cl₂ needed for all the Al: 0.37064 × (3 ÷ 2) = 0.55597 mol. We have only 0.49365 mol, so Cl₂ is limiting (even though its mass is larger).
  3. AlCl₃ from Cl₂: 0.49365 × (2 ÷ 3) = 0.32910 mol. M(AlCl₃) = 26.98 + 3(35.45) = 133.33 g/mol. m = 0.32910 × 133.33 = 43.88 g.
  4. Al used: 0.49365 × (2 ÷ 3) = 0.32910 mol = 8.879 g. Al left: 10.0 − 8.879 = 1.12 g.
  5. Check with conservation of mass: 35.0 g Cl₂ + 8.88 g Al used = 43.88 g AlCl₃. ✔

Cl₂ is limiting; 43.9 g of AlCl₃ forms; 1.12 g of Al is left over.

Worked example

Example 3: A precipitation reaction

Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq). A solution with 3.31 g of Pb(NO₃)₂ is mixed with one containing 5.00 g of KI. What mass of bright yellow PbI₂ precipitate can form?

  1. M: Pb(NO₃)₂ = 331.22, KI = 166.00, PbI₂ = 461.00 g/mol.
  2. n(Pb(NO₃)₂) = 3.31 ÷ 331.22 = 0.009993 mol. n(KI) = 5.00 ÷ 166.00 = 0.03012 mol.
  3. KI needed: 0.009993 × 2 = 0.01999 mol. We have 0.03012 mol, so KI is in excess and Pb(NO₃)₂ is limiting.
  4. n(PbI₂) = 0.009993 mol; m = 0.009993 × 461.00 = 4.607 g.

At most 4.61 g of PbI₂ can form.

Why industries use an excess on purpose

Chemical companies often add one reactant in excess on purpose:

  • to make sure an expensive or hazardous reactant is used up completely (the cheaper one is put in excess);
  • to make a reaction go faster or more completely.

For example, a furnace or engine is supplied with excess air so that the fuel burns completely. If oxygen becomes the limiting reagent, combustion is incomplete: carbon monoxide (a poisonous gas) and soot form instead of only CO₂ and water. You studied incomplete combustion in Grade 10; it is one reason every Ontario home with a fuel-burning appliance or an attached garage must have a working carbon monoxide alarm.

But excess has costs too: leftover reactant must be recovered, recycled or disposed of safely, which is part of the next lessons' story about yield and accuracy in industry.

Just for fun

What do you call the guest who leaves a party first and makes it end for everyone?

The limiting reagent. Everyone else was just in excess.

Hands-on activity

Cracker sandwiches: find the limiting reagent

You need: a box of crackers, a few slices of cheese cut into squares the size of a cracker (a grown-up can cut them), and a plate. Check for allergies first.

  1. Recipe A: 2 crackers + 1 cheese square → 1 cracker sandwich. Count out 9 crackers and 6 cheese squares. Predict the limiting reagent, the number of sandwiches and what is left over. Then build them and check.
  2. Recipe B (a "triple-decker"): 3 crackers + 2 cheese squares → 1 sandwich. Use 12 crackers and 7 cheese squares. Predict, then build.
  3. In each case, which ingredient did you have fewer of? Was that always the limiting one?
  4. Eat your products!

What to look for: Recipe A: 9 crackers make 4 sandwiches (4.5, but you cannot make half), 6 cheese could make 6, so crackers limit: 4 sandwiches, 1 cracker and 2 cheese left. Recipe B: 12 crackers could make 4, 7 cheese could make 3 (3.5), so cheese limits: 3 sandwiches, 3 crackers and 1 cheese left. In Recipe A you had fewer cheese squares, yet cheese was not limiting; the recipe ratio decides, just as the mole ratio does.

Your turn

1

2H₂ + O₂ → 2H₂O. 4.0 mol of H₂ is mixed with 4.0 mol of O₂. Find the limiting reagent, the amount of water formed and the excess left.

2

N₂ + 3H₂ → 2NH₃. 2.0 mol of N₂ and 3.0 mol of H₂ react. Which is limiting? What amount of NH₃ forms?

3

2Na(s) + Cl₂(g) → 2NaCl(s). 5.00 g of Na reacts with 5.00 g of Cl₂. Find the limiting reagent and the mass of NaCl formed.

4

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). 6.54 g of Zn is added to a solution containing 0.150 mol of HCl. Which is limiting? What mass of H₂ forms?

5

CH₄ + 2O₂ → CO₂ + 2H₂O. 16.0 g of CH₄ is burned with 48.0 g of O₂. Find the limiting reagent and the mass of CO₂ produced.

6

Why might a company add the cheaper reactant in excess?

More practice: warm-up

1

What is a limiting reagent?

2

What is an excess reagent?

3

Recipe: 2 bread + 1 cheese → 1 sandwich. You have 9 bread and 5 cheese. How many sandwiches can you make?

4

2Mg + O₂ → 2MgO. You have 2.0 mol of Mg and 2.0 mol of O₂. Which is limiting?

5

True or false: the reactant with the smaller mass is always the limiting reagent.

More practice: on your own

1

2Al + 3Cl₂ → 2AlCl₃. 2.70 g of Al reacts with 4.00 g of Cl₂. Find the limiting reagent and the mass of AlCl₃.

2

Fe₂O₃ + 2Al → Al₂O₃ + 2Fe. 50.0 g of Fe₂O₃ is mixed with 20.0 g of Al. Find the limiting reagent, the mass of iron formed and the mass of excess reagent left.

3

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq). 1.70 g of AgNO₃ (M = 169.88 g/mol) is mixed with 1.00 g of NaCl. What mass of AgCl forms?

4

A camping stove burns with a smoky yellow flame and leaves soot on the pot. In terms of limiting reagents, what is happening, and why is it a safety concern?

5

2H₂ + O₂ → 2H₂O. 3.00 g of H₂ is mixed with 16.0 g of O₂. Find the mass of water formed and the mass of excess reagent left.

Challenge

1

In a Haber-process reactor, 1.00 kg of N₂ reacts with 1.00 kg of H₂: N₂ + 3H₂ → 2NH₃. Find the limiting reagent, the maximum mass of NH₃ and the mass of excess reagent left.

2

What mass of oxygen is needed to burn exactly 10.0 g of methane, CH₄ + 2O₂ → CO₂ + 2H₂O, so that neither reactant is in excess?

3

A student mixes equal masses of zinc and sulfur to make zinc sulfide: Zn + S → ZnS. Without calculating, predict the limiting reagent and explain. Then check with 10.0 g of each.

Check your understanding

Quiz

1.

Which statement about the limiting reagent is true?

2.

2H₂ + O₂ → 2H₂O. 6 mol of H₂ and 2 mol of O₂ react. Which is limiting?

3.

2H₂ + O₂ → 2H₂O. 6 mol of H₂ and 2 mol of O₂ react. What amount of water forms?

4.

N₂ + 3H₂ → 2NH₃. 1.0 mol N₂ reacts with 6.0 mol H₂. How much H₂ is left over?

5.

Why do furnaces and engines run with excess air?

6.

Zn + S → ZnS. 0.20 mol Zn and 0.15 mol S react. What amount of ZnS forms?

For parents and teachers

Big idea: When reactants are not in the exact mole ratio, one runs out first (the limiting reagent) and decides how much product forms; the other is in excess. This lesson covers the limiting-reagent half of D2.6, using D3.4 mole ratios.

How to help: The cracker-sandwich activity is the best analogy there is; let your teen predict before building. Then ask: "In the chemistry problem, what are the crackers and what is the cheese?"

Common difficulties:

  • Choosing the reactant with the smaller mass, or fewer moles, without using the mole ratio.
  • Calculating product from the excess reagent (this gives too much product).
  • Forgetting to subtract to find the excess left over.
  • Not checking with conservation of mass, which catches many arithmetic slips.

Safety connection: the lesson explains why fuel-burning appliances need fresh air and why carbon monoxide alarms matter; this is a good moment to test the alarms at home together.