Limiting Reactant
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Concept 1. Limiting Reactant
C ONCEPT
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Limiting Reactant
Lesson Objectives
•
•
•
•
•
Identify a limiting reactant in your everyday life.
Identify limiting reactants in chemical reactions.
Identify excess reactants in chemical reactions.
Calculate the limiting reactant using the mole-mole ratios.
Calculate the products using the limiting reactant and the mass-mass ratios.
Introduction
Suppose you were in the business of building tricycles from the components of frames, handlebars, seats, and wheels.
After you completed constructing the tricycle, you boxed it up and shipped it. Suppose further that your component
supplies consisted of 100 frames, 120 sets of handlebars, 80 seats, and 300 wheels. What is the maximum number
of tricycles you could build and ship? With only a small amount of thinking, you probably came with 80 tricycles.
When you run out of seats, you can no longer complete and ship tricycles. The number of tricycles you could build
was limited by the number of seats you had available. The seats would be called the limiting component in your
tricycle construction. The other components, the ones you have more than enough of, are said to be in excess. If you
received a shipment of 100 more seats, could you build 100 more tricycles? Of course not, because one of the other
components would become the limiting component.
In the chemical reactions that you have been working with, you were given the amount of one of the substances
involved in the reaction and it was assumed that you had exactly the required amount of all the other reagents
(reacting chemicals). In this next section, you will be given the amounts of all the reagents but not in the correct
reacting amounts. Your task will be to determine which of the reactants is the limiting reactant and how much
product can be produced from that limiting reactant.
Limiting reactants determine the maximum amount of product that can be formed. Limiting reactants are all around
us. We are faced with them on a daily basis. When we work in the kitchen, baking, the limiting reagent may be the
ingredients that we have in the cupboard.
Sample Question: You are preparing lunch for a family reunion. You have 25 people attending and are making
sandwiches. You go to the store and buy 3 loaves of bread, 5 packages of chicken breast, and 2 bags of lettuce
leaves. Each bread bag has 18 slices in it, each package of chicken breast has 10 slices in it, and each bag of lettuce
has 30 leaves in it. In order to make one sandwich, you need two slices of bread, two slices of meat and two lettuce
leaves. What is your limiting “reactant?”
Solution:
18 slices × 3 loaves → 54 slices of bread → 27 possible sandwiches
10 slices × 5 packages → 50 slices of chicken breast → 25 possible sandwiches
2 bags × 30 leaves → 60 lettuce leaves → 30 possible sandwiches
Therefore, the limiting “reactant” is chicken breast, and you can make 25 sandwiches with lettuce and bread left
over. Hopefully everyone only takes one sandwich!
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Limiting Reactant: The One That’s Used Up in the Reaction
As we discussed in the last section, the limiting reactant is the one that is used up in the reaction. Often, when
we run experiments we want them to run to completion so we add one reactant in excess and have the other as the
limiting reactant. For example if we wanted to find out how much CO2(g) we would obtain if we combusted an
aspirin tablet we would use excess oxygen to make sure that the entire aspirin tablet reacted.
C9 H8 O4(s) + 9 O2(g) → 9 CO2(g) + 4 H2 O(L)
The limiting reagent is the aspirin tablet because it is the one used up in the experiment. The other reagent, the
oxygen, is in excess.
Let’s look at another example. A common reaction is when you take a piece of aluminum foil and place it into a
solution of copper (II) chloride. Rather quickly, there’s lots of heat and gas given off. When the reaction to goes to
completion, the blue color of the copper (II) chloride solution fades and a brownish solid is produced, that is, copper.
What is the maximum number of grams of copper that can be produced when 53.96 grams of aluminum are reacted
with 134.4 grams of copper (II) chloride?
The reaction is:
2 Al(s)
+
3 CuCl2(aq)
2 moles
3 moles
53.96 g
134.44 g
→
3 Cu(s)
3 moles
+
2 AlCl3(aq)
2 moles
Since 53.96 g of Al is 2 moles; and, 134.44 grams of CuCl2 is 3 moles, we know that both reactants will be
completely used up and 3 moles of Cu, or 187.3 grams, will be produced.
But, instead of having 53.96 grams of Al, suppose we only have 10.0 grams , what is the maximum number of grams
of copper that can be produced? Assuming we still have the 134.44 grams of CuCl2 , then it must be obvious that the
Al is the limiting reactant and only 35.5 grams of Cu will be produced. Not all reactions are this obvious.
Look at another example. Assume that we have 72.0 g of water reacting with 190. g of TiCl4 . The balanced equation
is:
TiCl4(s) + 2 H2 O(L) → TiO2(s) + 4 HCl(aq)
We can see from the balanced equation that one mole of TiCl4 reacts with two moles of water. The 190. g of TiCl4
is exactly one mole. Therefore, this TiCl4 will react with two moles or 36.0 g of water. But in this reaction, we are
given 72.0 g of water. Therefore, the water is in excess and 36.0 g of water will be left over when the reaction is
complete.
Sample Question
You are given 34.0 g of NH3 and 36.5 g of HCl which react according to the following balanced equation. Determine
the limiting reactant and the reactant in excess.
NH3(g) + HCl(aq) → NH4Cl(s)
According to the balanced equation, one mole of NH3 reacts with one mole of HCl. The 34.0 g of NH3 is two moles
and the 36.5 g of HCl is one mole. Since they react with a ratio of one-to-one, one mole of HCl will react with one
mole of NH3 and one mole of NH3 will be left over. HCl is the limiting reactant and NH3 is in excess.
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Concept 1. Limiting Reactant
Limiting Reactant Determined Using the Mole-Mole Ratio
We can use a stepwise strategy as we did in earlier mathematical calculations to determine the limiting reactant.
Remember that with chemical equations, the key to the entire chemical equation is the mole. The balancing
coefficients represent the number of moles in the chemical equation. Look at the equation below.
Al4C3 + 12 H2 O → 4 Al(OH)3 + 3CH4
We can say that one mole of aluminum carbide reacts with twelve moles of water to produce four moles of aluminum
hydroxide and three moles of methane. Notice how the amounts of each substance relate to each other in terms of
the number of moles. This is the reason we learned to balance equations. Once we have our balanced chemical
equation, we can determine such key features as the limiting reagent(s), and the excess reagent(s) using a stepwise
strategy.
Start with a balanced equation and then follow these steps for finding the limiting reagent. Step 1: Determine the
number of moles given of each reactant.
Step 2: Determine the moles of reactant A that are required to consume all of reactant B. This is done by multiplying
the given moles of B by the molar ratio A/B.
Step 3: Compare the necessary moles of reactant A found in step 2 with actual amount given in step 1. If the amount
of A in step 2 is greater than that in step 1, A is the limiting reactant. If the amount of A in step 1 is greater than in
step 2, B is the limiting reactant.
Let’s try an example:
Andrew was working in the lab on the reaction between hydrochloric acid and iron(II) sulfide. He mixed a solution
containing 25.0 g of HCl and one containing 25.0 g of iron (II) sulfide. What is the limiting reagent and what reagent
was in excess?
The balanced equation is:FeS + 2 HCl → H2 S + FeCl2
Step 1: Determine the number of moles of each reactant in the chemical equation.
mols FeS =
25.0 g
25.0 g
= 0.284 mol FeS mols HCl =
= 0.686 mol HCl
87.9 g/mol
36.5 g/mol
Step 2: Determine the moles of FeS that are required to consume all of the HCl.
mol FeS
mols FeS required to react with all of the HCl = (0.686 mol HCl) 21 mol
HCl = 0.343 mols FeS
Step 3: Compare moles of FeS found in step 2 with actual amount given in step 1. If the amount in step 2 is greater
than in step 1, FeS is limiting reagent. If the amount in step 2 is less than in step 1, FeS is in excess.
We know that we have 0.284 mol of FeS (Step 1)
We know that we need 0.343 mol of FeS (Step 2) to use up all of the HCl.
Therefore step 2 >step1, and FeS is the limiting reagent. This means that HCl is the species in excess.
Let’s try another one:
A student mixed 5.0 g of carbon with 23.0 g iron (III) oxide. What is the limiting reactant and what reactant was in
excess? The reaction she used is below:
2 Fe2 O3 + 3 C → 4 Fe + 3 CO2
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Step 1: Determine the number of moles of each reactant in the given quantities.
mols C =
5.0 g
23.0 g
= 0.417 mol C mols Fe2 O3 =
= 0.144 mol Fe2 O3
12.0 g/mol
159.7 g/mol
Step 2: Determine the moles of Fe2 O3 that are required to consume all of the carbon.
Fe2 O3
mols ofFe2 O3 required to react with all the carbon = (0.143 mol C) 2 mols
= 0.277 mols Fe2 O3
3 mols C
Step 3: Compare moles of Fe2 O3 found in step 2 with actual amount given in step 1. If step 2 >step 1, Fe2 O3 is
limiting reagent. If step 2 >step 1, Fe2 O3 is in excess.
We know that we have 0.144 mol of Fe2 O3 (Step 1)
We know that we need 0.275 mol of Fe2 O3 (Step 2) to use up the entire C.
Therefore Step 2 >Step1, and Fe2 O3 is the limiting reagent. This means that C is the species in excess.
Sample Problem: While working in the lab, Brenda decided to add 5.0 g of aluminum chloride to 7.0 g of silver
nitrate in order to work on her double replacement reactions. Her first goal was to determine the limiting reagent.
Can you help her determine which reactant is limiting and which is in excess? Her reaction is listed below.
AlCl3 + 3 AgNO3 → 3 AgCl + Al(NO3 )3
Solution:
Step 1: Determine the number of moles of each reactant in the given quantities.
mols AlCl3 =
5.0 g
7.0 g
= 0.037 mol AlCl3 mols AgNO3 =
= 0.041 mol AgNO3
133 g/mol
170. g/mol
Step 2: Determine the moles of AlCl_3 that are required to consume all of the AgNO3 .
mol AlCl3
mols AlCl3 required to react with all the AgNO3 = (0.041 mol AgNO3 ) 31mol
AgNO3 = 0.014 mol AlCl3
Step 3: Compare moles of AlCl3 found in step 2 with actual amount given in step 1. If step 2 >step 1, AlCl3 is
limiting reagent. If step 2 >step 1, AlCl3 is in excess.
We know that we have 0.037 mol of AlCl3 (Step 1)
We know that we need 0.014 mol of AlCl3 (Step 2) to use up the entire AgNO3 .
Therefore Step 2 >Step1, and AlCl3 is in excess. This means that AgNO3 is the limiting reagent.
After Limiting Reactant Determined, It’s a Mass-Mass Calculation
Once we have determined the limiting reactant, we still need to determine the amount of product the limiting reactant
will produce. We might even want to know how much of the other reactant we are using and how much is left over.
We want these answers in grams. Using one of the examples above, we now want to determine the mass of iron
produced. We simply have to add two additional steps. Here is the equation and the moles of limiting reactant we
determined.
2 Fe2 O3 + 3 C → 4 Fe + 3 CO2
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Concept 1. Limiting Reactant
Fe2 O3 is the limiting reagent and 23.0 g was placed in the container which is equivalent to 0.144 mol.
Step 4: Convert moles of limiting reactant to moles of product (using molar ratios).
Step 5: Convert moles of product to mass ( usingmass = moles × molar mass).
Step 4:
mols Fe = (0.144 mol Fe2 O3
4 mols Fe
2 mols Fe2 O3
= 0.28 mol Fe
Step 5:
mass Fe = 0.28 mol Fe × 55.85 g/mol
mass Fe = 16 g Fe
Therefore the mass of Fe formed from this reaction would be 16 g.
Now, let’s do one problem from the top:
Sample Problem #1
Magnesium chloride is used as a fireproofing agent but surprisingly it is also a coagulant in tofu – imagine that! Jack
was working in the lab one day and mixed 4.56 g of magnesium chloride with 8.25 g of water. The reaction is given
below. What mass of magnesium oxide was formed? (Hint: Remember to balance the equation first!)
MgCl2 + H2 O → MgO + HCl
Solution:
MgCl2 + H2 O → MgO + 2 HCl
mols MgCl2 =
4.56 g
= 0.0479 mol MgCl2
95.2 g/mol
mols H2 O =
8.25 g
= 0.458 mol H2 O
18.0 g/mol
Step 2: Determine the moles of MgCl2 that are required to consume all of the H2 O.
mols MgCl required to consume all the H O = (0.458 mol H O) 1 mol MgCl2 = 0.458 mol MgCl
2
2
2
1 mol H2 O
2
Step 3: Compare moles of MgCl2 found in step 2 with actual amount given in step 1. If step 2 >step 1, MgCl2 is
limiting reagent. If step 2 <step 1, MgCl2 is in excess.
We know that we have 0.0479 mol of MgCl2 (Step 1)
We know that we need 0.458 mol of MgCl2 (Step 2) to use up the entire H2 O.
Since step 2 >step 1, MgCl2 is the limiting reactant. This means that H2 O is in excess.
Step 4:
1 mol MgO
mols MgO = (0.0479 mol MgCl2 )
1 mol MgCl2
= 0.0479 mol MgO
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Step 5:
mass MgO = 0.0479 mol MgO × 40.31 g/mol = 1.93 g MgO
Therefore the mass of MgO formed from this reaction would be 1.93 g.
Sample Problem #2: In the following reaction, carried out in a laboratory, 12.8 g of ammonia and 14.5 g of oxygen
were allowed to react. How many grams of nitrogen monoxide were formed?
4 NH3 + 5 O2 → 4 NO + 6 H2 O (balanced)
Solution:
Step 1: Convert the given quantities to moles.
mols NH3 =
12.8 g
= 0.752 mol NH3
17.0 g/mol
mols O2 =
14.5 g
= 0.453 mol O2
32.0 g/mol
Step 2: Determine the moles of NH3 that are required to consume all the O2 .
NH3
mols NH3 required to consume all the O2 = (0.453 mol O2 ) 45mol
= 0.362 mol NH3
mol O2
Step 3: Compare moles of NH3 found in step 2 with actual amount given in step 1. If step 2 >step 1, NH3 is limiting
reagent. If step 2 <step 1, NH3 is in excess.
We know that we have 0.752 mol of NH3 (Step 1)
We know that we need 0.362 mol of NH3 (Step 2) to use up the entire O2 .
Since step 2 <step1, NH3 is in excess. This means that O2 is limiting reagent.
Step 4:
mols NO = (0.453 mol O2
4 mols NO
5 mols O2
= 0.362 mol NO
Step 5:
mass NO = 0.362 mol NO × 30.01g/mol = 10.9g NO
Therefore the mass of NO formed from this reaction would be 10.9 g.
Lesson Summary
• The limiting reagent is the one that will limit the amount of product that is formed. The reactant that is not
the limiting reagent is considered to be “in excess.” The calculations to determine the limiting reagent are
determined using moles.
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Concept 1. Limiting Reactant
Review Questions
1. Consider the balanced reaction: 2 Al + 6 HBr → 2 AlBr3 + 3 H2 .
a. When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed?
b. What is the limiting reactant?
c. For the reactant in excess, how many moles are left over at the end of the reaction?
2. (a) Write the balanced equation for this reaction: copper (II) chloride reacts with sodium nitrate to form
copper (II) nitrate and sodium chloride. (b) If 15.0 grams of copper (II) chloride react with 20.0 grams of
sodium nitrate, how much sodium chloride can be formed? (c) What is the limiting reactant for this reaction?
(d) How much of the non-limiting reactant will be left over in this reaction?
Supplemental Links
http://en.wikipedia.org
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