MathConnections
CH 3 CHCH 2 CH3CH2CH2CH2
4.3 Limiting Reactants
To put that in more familiar units, use the fact that a pound is approximately 454 g.
Our calculation then says that burning a gallon of gasoline will produce 18 pounds of carbon dioxide. Although we made a few simplifying assumptions, our result should certainly be a reasonable estimate.
So in any calculation related to fuel consumption or exhaust emissions, it would be reasonable to assume that gasoline would always be the limiting reactant.
In other circumstances, the identity of the limiting reactant may not be so obvi- ous. In a laboratory experiment, we might simply mix conveniently available quanti- ties of substances to begin to study their reaction. Determining the limiting reactant requires comparing the amount of each reactant present. As always, we should keep in mind that balanced chemical equations tell us the ratios in which moles of one substance react with moles of another. So we cannot compare masses to determine the limiting reactant; we must do the comparison using the relevant mole ratios. Example Problems 4.4 and 4.5 show how we can determine which substance is in excess and which is the limiting reactant.
E X A M P L E P RO B L E M 4 . 4
A solution of hydrochloric acid contains 5.22 g of HCl. When it is allowed to react with 3.25 g of solid K2CO3, the products are KCl, CO2, and H2O. Which reactant is in excess?
Strategy As for any reaction stoichiometry problem, we should start with a balanced equation. One way to proceed from there is to calculate the amount of one reactant that would combine with the given amount of the second reactant. Comparing that with the amount actually available will reveal the limiting reactant.
Solution We know the reactants and products, so we can write a skeleton equation easily:
HCl + K2CO3: KCl + CO2+ H2O
Balancing this gives us the chemical equation we need for our stoichiometry calculation:
2 HCl + K2CO3: 2 KCl + CO2+ H2O
Let’s use the quantity of HCl given and calculate the amount of K2CO3 that would react completely with it.
5.22 g HCl × 1 mol HCl
——————
36.46 g HCl × 1 mol K2CO3
———————
2 mol HCl × 138.2 g K2CO3
———————
1 mol K2CO3 = 9.89 g K2CO3 So the given quantity of HCl (5.22 g) requires 9.89 g K2CO3, but we have only 3.25 g of K2CO3 available. The reaction will stop once all of the K2CO3 is consumed. K2CO3 is the limiting reactant, and HCl is in excess.
Analyze Your Answer It’s often hard to gain an intuitive feel for limiting reac- tants, so that is why it is important to learn this type of calculation. There are two factors that determine the limiting reactant—the relative molar masses and the stoi- chiometric mole ratio. In this case, the molar mass of K2CO3 is about three times larger than that of HCl, so it is not surprising that similar masses of the two reactants would lead to K2CO3 as the limiting reactant.
Check Your Understanding Ammonia is widely used in the production of fer- tilizers and is also a precursor to a number of important polymers. So the formation of ammonia from its elements is one of the most important industrial chemical reactions:
N2+ 3 H2: 2 NH3
This process is usually carried out using an excess of hydrogen. If a reaction vessel contains 1.5 × 103 moles of N2, how much hydrogen would be needed to ensure that it will be in excess?
There are a variety of possible strategies for identifying a limiting reactant.
There are a variety of possible strategies for identifying a limiting reactant.
E X A M P L E P RO B L E M 4 . 5
In Example Problem 4.3, we used the reaction that produces P4S3, one of the reactants in the combustion of a match:
8 P4+ 3 S8: 8 P4S3
If 28.2 g of P4 is allowed to react with 18.3 g of S8, which is the limiting reactant?
Strategy We can choose either reactant and determine how much of the other reactant is required to consume it entirely. Comparing the amount calculated with the amount given in the problem will let us determine which reactant is limiting.
Solution Let’s begin with P4: 28.2 g P4× 1 mol P4
—————
123.9 g P4 × 3 mol S8
—————
8 mol P4 × 256.5 g S8
—————
1 mol S8 = 21.9 g S8
So, 28.2 g of P4 requires 21.9 g of S8 to react completely. We have only 18.3 g of S8, so there is not enough S8 to react with all of the P4. Therefore, S8 is the limiting reactant.
Check Your Understanding To convince yourself that you can start with either reactant, redo the example starting with the quantity of S8 given.
The determination of the limiting reactant is typically just a piece of a larger puzzle.
In most limiting reactant stoichiometry problems, the real goal is to determine how much product could be formed from a particular reactant mixture. Perhaps the fi rst critical step in such a problem is to correctly recognize the fact that it is a limiting reactant situation.
Whenever a particular amount of more than one reactant is specifi ed in a stoichiometry problem, it should tell us to begin by fi nding the limiting reactant. Once we know which substance runs out fi rst, we are left with an ordinary stoichiometry calculation to solve the problem. Example Problem 4.6 shows how this type of problem can be approached.
E X A M P L E P RO B L E M 4 . 6
MTBE (methyl tert-butyl ether) has been used as an additive in gasoline. The com pound is produced by reacting methanol and isobutene, according to the following equation:
CH3OH + (CH3)2C"CH2: (CH3)3COCH3
Methanol Isobutene MTBE
If 45.0 kg of methanol is allowed to react with 70.0 kg of isobutene, what is the maximum mass of MTBE that can be obtained?
Strategy We are given amounts of two different reactants here (methanol and iso- butene). So we should realize this is a potential limiting reactant situation. We can identify the limiting reactant as in the previous examples: choose one reactant and fi nd the amount of the other that would be needed to react with it. Comparing that result with the given amount, we can determine which of the reactants is limiting. Once we have done that, it will be straightforward to calculate the expected amount of product.
Solution We’ll need the molar masses of all three compounds. Taking care to inter- pret the parentheses in the molecular formulas correctly, we fi nd the following:
Methanol (CH3OH): 32.042 g/mol Isobutene ((CH3)2C"CH2): 56.107 g/mol MTBE ((CH3)3COCH3): 88.149 g/mol We will discuss the use of such additives
in the closing section of this chapter.
We will discuss the use of such additives in the closing section of this chapter.
To identify the limiting reactant, we can calculate the mass of isobutene that would react with 45.0 kg (or 45,000 g) of methanol.
45,000 g methanol × ————————— 1 mol methanol
32.042 g methanol × 1 mol isobutene ————————
1 mol methanol × 56.107 g isobutene —————————
1 mol isobutene
= 7.88 × 104 g isobutene = 78.8 kg isobutene This tells us that to use 45.0 kg of methanol, we need at least 78.8 kg of isobutene.
But we have only 70.0 kg of isobutene. So the amount of isobutene available will determine the amount of MTBE that can be formed. Isobutene is the limiting reactant and methanol is in excess.
Now the problem is reduced to a simpler stoichiometry calculation starting from the mass of isobutene available.
70,000 g isobutene × 1 mol isobutene
—————————
56.107 g isobutene × 1 mol MTBE
————————
1 mol isobutene × 88.149 g MTBE
————————
1 mol MTBE
= 1.10 × 105 g MTBE = 1.10 3 102 kg MTBE Analyze Your Answer This value is larger than the original amount of either reactant. Does this make sense? If we look closely at the reaction, it is a synthesis where the two molecules become one. If our answer had been greater than the sum of the two reactant masses, we would have cause for concern. But this value is certainly plausible.
Discussion This problem combines two types of calculations that we’ve seen in earlier examples. The fi rst step is identifying the limiting reactant, and the second is the actual stoichiometry calculation. Note that here isobutene was the limiting reac- tant even though the available mass of isobutene was greater than that of methanol.
This is a reminder that reaction stoichiometry calculations should always be handled in terms of moles, not masses.
Check Your Understanding The compound diborane, B2H6, was once proposed for use as a rocket propellant. It can be produced by the following reaction
3 LiBH4+ BF3: 2 B2H6+ 3 LiF
If 24.6 g of LiBH4 is combined with 62.4 g of BF3, what mass of diborane can be formed?
The concept of limiting reactants has some important implications for engineering applications where chemical reactions are involved. We’ve already mentioned that a good design for a chemical process would ensure that a scarce or expensive substance would be the limiting reactant, so that none of it would go to waste. A rather different situation exists for the design of rocket engines. Here the total mass of the rocket is a major con- sideration, so the aim would likely be to use the minimum mass of fuel that could deliver the needed thrust. Many rocket engines rely on so-called two-component fuel mixtures, where energy is released as two compounds are allowed to react. The optimal design for such a rocket would generally ensure that the two compounds are present in a stoichio- metric mixture, so that neither is in excess. The following example illustrates how such a mixture might be chosen.
E X A M P L E P RO B L E M 4 . 7
The solid fuel booster rockets of the space shuttle are based on the following reaction between ammonium perchlorate and aluminum:
3 NH4ClO4(s) + 3 Al(s) : Al2O3(s) + AlCl3(g) + 3 NO(g) + 6 H2O(g)
If either reactant is in excess, unnecessary mass will be added to the shuttle, so a stoichiometric mixture is desired. What mass of each reactant should be used for every kilogram of the fuel mixture?
Strategy We want to ensure that there will be no excess of either reactant. Because we are asked for the composition “for every kilogram” of fuel, we might start by assuming the total mass of fuel will be 1000 g. We can write that as an equation with two unknowns:
mNH4ClO4+ mAl= 1000 g
Because we have two unknowns, we will need a second relationship between the two masses. The balanced chemical equation shows us that the mole ratio between the two reactants should be 3:3, or 1:1. We can use molar masses to convert that to a ratio in terms of masses and then use that ratio as a second equation to solve the problem.
Solution The molar mass of NH4ClO4 is 117.49 g/mol, and that of Al is 26.98 g/mol.
To ensure that we have a stoichiometric mixture, we need equal molar amounts of each reactant. So if we were to use 117.49 g (1 mole) of NH4ClO4, we would need 26.98 g (also 1 mole) of Al. We can write this as an equation:
mNH4ClO4
————— mAl = 117.49 g
—————
26.98 g = 4.355
And although our actual amounts will need to be larger than those, that ratio of masses will still be right. So now we have two equations for the two unknown masses, and we can solve by doing a little bit of algebra. Our two equations are
mNH4ClO4+ mAl= 1000 g mNH4ClO4= 4.355 mAl
Substituting the right-hand side of the second equation in the fi rst equation gives:
4.355 mAl + mAl = 1000 g
Simplifying that and solving, we get the mass of aluminum needed for each kilogram of the fuel mixture.
5.355 mAl = 1000 g mAl = 1000 g
————
5.355 = 186.8 g Al Now the mass of NH4ClO4 is easy to obtain.
mNH4ClO4= 1000 g − mAl
= 1000 g − 186.8 g = 813.2 g
So each kilogram of fuel should comprise 186.8 g Al and 813.2 g NH4ClO4.
Discussion The mass of fuel required for a space shuttle engine would clearly be much larger than a kilogram, but this result could easily be scaled up to the needed mass. As for most problems, there are a variety of other approaches that could be used to solve this. The particular strategy that we’ve used is based on the general idea that if we have two unknown quantities (like the two masses here), we need to fi nd two separate relationships between them to solve the problem.
Check Your Understanding A fuel mixture comprising 600.0 g of NH4ClO4 and 400.0 g of Al is used in a test of a laboratory scale mock-up of a shuttle engine.
When the engine stops burning, some unused fuel remains. What substance is this unburned fuel, and what mass of it should be found?