06.2 - Chemical equations and reaction stoichiometry · Chemical Equations HH H O O O C O H H H H H C O O O reaction stoichiometry: the quantitative relationships among substances

Chemical equations and reaction stoichiometry

Chemical Equations

H H HOO

O C O

H

H

H

HHC

OOO

O

reaction stoichiometry: the quantitative relationships among substances as they participate in chemical reactions.

Chemical equations represent a very precise, yet a very versatile, language that describes chemical changes

Chemical Equations

Chemical reactions always involve changing one or more substances into one or more different substances. In other words, chemical reactions rearrange atoms or ions to form other substances.

Chemical equations are used to describe chemical reactions, and they show

•The substances that react, called reactants; • The substances formed, called products•The relative amounts of the substances involved.

Chemical Equations

We write the reactants to the left of an arrow and the products to the right of the arrow. The numbers placed in front of compounds in a chemical equation are called coefficients and represent the number of molecules of each reactant or product needed to balance the equation.

CH4 1 2O2 88n CO2 1 2H2O

reactants products

Subscripts indicate the number of atomsof each element in the compound or element and CANNOT be changed

when balancing an equation.

Coefficients indicate the amount of eachcompound or element and can be

changed to balance the equation; omitted coefficients are assumed to be 1.

Balancing Chemical Equations

The combustion (burning) of natural gas is a reaction used to heat buildings and cook food. The principal natural gas is methane, CH4. The equation that describes the reaction of methane with excess oxygen is:

The equation tells us that methane reacts with oxygen to produce carbon dioxide, CO2, and water. More specifically, for every CH4 molecule (or mole) that reacts, two molecules (or moles) of O2 also react, and that one CO2 molecule (or mole) and two H2O molecules (or mole) are formed.

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3 - 1 C H E M I C A L E Q U A T I O N S 83

How do we interpret this equation? In the simplest terms, it tells us that methane reacts with oxygen to produce carbon dioxide, CO2, and water. More specifically, it says that for every CH4 molecule that reacts, two molecules of O2 also react, and that one CO2 mole-cule and two H2O molecules are formed. That is,

CH4 1 2O2 h CO2 1 2H2O 1 molecule 2 molecules 1 molecule 2 molecules

This description of the reaction of CH4 with O2 is based on experimental observations. Many experiments have shown that when one CH4 molecule reacts with two O2 mole-cules, one CO2 molecule and two H2O molecules are formed. Chemical equations are based on experimental observations. Special conditions required for some reactions are indicated by a notation above or below the arrow. Figure 3-1 is a pictorial representation of the rearrangement of atoms described by this equation.

In Section 1-1, we learned the Law of Conservation of Matter: there is no detectable change in the quantity of matter during an ordinary chemical reaction. This guiding principle is the basis for “balancing” chemical equations and for performing calculations based on those equations. Because matter is neither created nor destroyed during a chemical reaction,

a balanced chemical equation must always include the same number of each kind of atom on both sides of the equation.

Chemists usually write equations with the smallest possible whole-number coefficients.Before we attempt to balance an equation, all substances must be represented by

formulas that describe them as they exist. For instance, we must write H2 to represent diatomic hydrogen molecules—not H, which represents individual hydrogen atoms, which are unstable and do not exist by themselves under normal conditions. Once the correct formulas are written, the subscripts in the formulas may not be changed. Different sub-scripts in formulas specify different substances, so changing the formulas would mean that the equation would no longer describe the same reaction.

heat ▶ The arrow may be read “yields.” The capital Greek letter delta (D) is often used in place of the word “heat.”

Figure 3-1 Three representations of the reaction of methane with oxygen to form carbon dioxide and water. Chemical bonds are broken and new ones are formed in each representation. Part (a) illustrates the reaction using ball-and-stick models, (b) uses chemical formulas, and (c) uses space-filling models.

H

H +

+

+O O

O O

O O

O

OOOH

H

HHH

H H

O

H H

HH

O

HH

C

CC

C

H

CH4 2O2 +

+ +

CO2 2H2O

reactants products

1C, 4H, 4O atoms 1C, 4H, 4O atoms

O O

O O

A

C

B

10663_03_ch03_p081-114.indd 83 12/7/12 1:51 PM

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The arrow may be read “yields.” The capital Greek letter delta (Δ) is often

used in place of the word “heat.”


Following the Law of Conservation of Matter, there is no detectable change in the quantity of matter during an ordinary chemical reaction. This guiding principle is the basis for “balancing” chemical equations and for performing calculations based on those equations.

Because matter is neither created nor destroyed during a chemical reaction, a balanced chemical equation must always include the same number of each kind of atom on both sides of the equation.

When balancing chemical equations, always write equations with the smallest possible whole-number coefficients.



How do we interpret this equation? In the simplest terms, it tells us that methane reacts with oxygen to produce carbon dioxide, CO2, and water. More specifically, it says that for every CH4 molecule that reacts, two molecules of O2 also react, and that one CO2 mole-cule and two H2O molecules are formed. That is,

CH4 1 2O2 h CO2 1 2H2O 1 molecule 2 molecules 1 molecule 2 molecules

This description of the reaction of CH4 with O2 is based on experimental observations. Many experiments have shown that when one CH4 molecule reacts with two O2 mole-cules, one CO2 molecule and two H2O molecules are formed. Chemical equations are based on experimental observations. Special conditions required for some reactions are indicated by a notation above or below the arrow. Figure 3-1 is a pictorial representation of the rearrangement of atoms described by this equation.

In Section 1-1, we learned the Law of Conservation of Matter: there is no detectable change in the quantity of matter during an ordinary chemical reaction. This guiding principle is the basis for “balancing” chemical equations and for performing calculations based on those equations. Because matter is neither created nor destroyed during a chemical reaction,

a balanced chemical equation must always include the same number of each kind of atom on both sides of the equation.

Chemists usually write equations with the smallest possible whole-number coefficients.Before we attempt to balance an equation, all substances must be represented by

formulas that describe them as they exist. For instance, we must write H2 to represent diatomic hydrogen molecules—not H, which represents individual hydrogen atoms, which are unstable and do not exist by themselves under normal conditions. Once the correct formulas are written, the subscripts in the formulas may not be changed. Different sub-scripts in formulas specify different substances, so changing the formulas would mean that the equation would no longer describe the same reaction.

heat ▶ The arrow may be read “yields.” The capital Greek letter delta (D) is often used in place of the word “heat.”

Figure 3-1 Three representations of the reaction of methane with oxygen to form carbon dioxide and water. Chemical bonds are broken and new ones are formed in each representation. Part (a) illustrates the reaction using ball-and-stick models, (b) uses chemical formulas, and (c) uses space-filling models.

H

H +

+

+O O

O O

O O

O

OOOH

H

HHH

H H

O

H H

HH

O

HH

C

CC

C

H

CH4 2O2 +

+ +

CO2 2H2O

reactants products

1C, 4H, 4O atoms 1C, 4H, 4O atoms

O O

O O

A

C

B

10663_03_ch03_p081-114.indd 83 12/7/12 1:51 PM



Dimethyl ether, C2H6O, burns in an excess of oxygen to give carbon dioxide and water.Let’s balance the equation for this reaction. In unbalanced form, the equation is:

C2H6O 1 O2 h CO2 1 H2O

Carbon appears in only one compound on each side, and the same is true for hydrogen. We begin by balancing these elements. The subscripts on the carbon and hydrogen atoms in C2H6O guide us in assigning the coefficients on the product side:

C2H6O ! O2 88n 2CO2 ! 3H2O

2C, 6H, 3Oatom count:

3 " 2 # 6

2C, 6H, 7O

reactants products

Remember to multiply all the atoms in a compound, including the subscript numbers, by the coefficient. Do

not change the formulas of the elements or compounds!


Now there is a different total number of O atoms on each side: 3 on the reactant side and 7 on the product side. We now need to balance the oxygen count by adding 4 more oxygen atoms to the reactant side. This is done by adding 2 more O2 molecules (total of 4 oxygen atoms) to the one already present giving a total of 3O2 molecules on the reactant side:

C2H6O ! O2 88n 2CO2 ! 3H2O


3 " 2 # 6

2C, 6H, 7O

C2H6O ! 3O2 88n 2CO2 ! 3H2O

C2H6O ! O2 88n 2CO2 ! 3H2O

C2H6O ! 3O2 88n 2CO2 ! 3H2O

2C, 6H, 7Ofinal atom count:

4 oxygen atoms " 2O2

2C, 6H, 7O

!


C2H6O ! 3O2 88n 2CO2 ! 3H2O

When we think we have finished the balancing, we should always do a complete check that the total number of atoms for each of the elements on the reactant side matches that found on the product side of the equation.

reactants products

final atom count: 2C, 6H, 7O

3O2 3H2O2CO2C2H6O

2C, 6H, 7O

+ +


Let’s examine a reaction that produces a fractional coefficient. Butane, C4H10, is used inmany lighters. The combustion reaction with O2 is initially written in unbalanced form as:

Carbon only appears in one compound on each side and the same is true for hydrogen. We begin by balancing these elements. The subscripts on the carbon and hydrogen atoms in C4H10 guide us in assigning the coefficients on the product side:

C4H10 1 O2 h CO2 1 H2O


C4H10 ! O2 88n 4CO2 ! 5H2O


5 " 2 # 10

4C, 10H, 13O

reactants products

There are different total numbers of O atoms on each side: 2 on the reactant and 13 on the product side. We need to balance the oxygen count by adding 11 oxygen atoms to the reactant side. This is done by adding 1½ 5 5.5 O2 molecules to the one that is currently there, giving a total of 6½ or 6.5 O2 molecules on the reactant side:

C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O

4C, 10H, 13Ofinal atom count: 4C, 10H, 13O

Although this is considered a properly balanced equation, many chemists do not ordinar-ily use fractional coefficients in chemical equations. To convert the fractional coefficient into an integer value, one simply multiplies the entire balanced equation by the proper integer value to convert the fractional value into the smallest integer value. In this case, one would multiply the entire equation by a factor of 2 to produce the final balanced equa-tion with all integer coefficients:

2C4H10 ! 13 O2 88n 8CO2 ! 10H2O

2 [ C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O ]


Let’s generate the balanced equation for the reaction of aluminum metal with hydrogen chloride to produce aluminum chloride and hydrogen. The unbalanced “equation” is

Al 1 HCl h AlCl3 1 H2

As it now stands, the “equation” does not satisfy the Law of Conservation of Matter because there are two H atoms in the H2 molecule and three Cl atoms in one formula unit of AlCl3 (product side), but only one H atom and one Cl atom in the HCl molecule (reac-tant side).

Let’s first balance chlorine by putting a coefficient of 3 in front of HCl.

Al ! 3HCl 88n AlCl3 ! H2

Al, 3H, 3Clatom count: Al, 2H, 3Cl

reactants products

Now there are 3H on the left and 2H on the right. We can add hydrogen only as two atoms at a time (as H2) on the product side. So we need to find the least common multiple of 3 and 2, which is 6, in order to balance the H atoms. To get a total of 6H atoms on each side, we multiply the 3HCl by 2 and the lone H2 by 3. This now gives

Al ! 6HCl 88n AlCl3 ! 3H2

"3Al ! 3HCl 88n AlCl3 ! H2


"2

▶ If you are going to express a coefficient as a fractional value it is best to list it as a decimal number, e.g., 6.5 instead of 61/2 or 13/2.


10663_03_ch03_p081-114.indd 85 12/7/12 1:51 PM


There are different total numbers of O atoms on each side: 2 on the reactant and 13 on the product side.



C4H10 ! O2 88n 4CO2 ! 5H2O


5 " 2 # 10

4C, 10H, 13O

reactants products


C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O



2C4H10 ! 13 O2 88n 8CO2 ! 10H2O

2 [ C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O ]








reactants products





"2



10663_03_ch03_p081-114.indd 85 12/7/12 1:51 PM


We need to balance the oxygen count by adding 11 oxygen atoms to the reactant side. This is done by adding a total of 6½ or 6.5 O2 molecules on the reactant side:


C4H10 ! O2 88n 4CO2 ! 5H2O


5 " 2 # 10

4C, 10H, 13O

reactants products


C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O



2C4H10 ! 13 O2 88n 8CO2 ! 10H2O

2 [ C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O ]








reactants products





"2



10663_03_ch03_p081-114.indd 85 12/7/12 1:51 PM


Although this is considered a properly balanced equation, chemists do not use fractional coefficients in chemical equations.



C4H10 ! O2 88n 4CO2 ! 5H2O


5 " 2 # 10

4C, 10H, 13O

reactants products


C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O



2C4H10 ! 13 O2 88n 8CO2 ! 10H2O

2 [ C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O ]








reactants products





"2



10663_03_ch03_p081-114.indd 85 12/7/12 1:51 PM


To convert the fractional coefficient into an integer value, multiply the entire balanced equation by the proper integer value to convert the fractional value into the smallest integer value. In this case, multiply the entire equation by a factor of 2 to produce the final balanced equation with all integer coefficients:


C4H10 ! O2 88n 4CO2 ! 5H2O


5 " 2 # 10

4C, 10H, 13O

reactants products


C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O



2C4H10 ! 13 O2 88n 8CO2 ! 10H2O

2 [ C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O ]








reactants products





"2



10663_03_ch03_p081-114.indd 85 12/7/12 1:51 PM


Problem-Solving Tip for Balancing Chemical Equations

There is no one best place to start when balancing a chemical equation, but

the following suggestions might be helpful:

1. Look for elements that appear in only one place on each side of

the equation (in only one reactant and in only one product), and

balance those elements first.

2. A good starting point is usually to pick a pair of compounds with a

common element. Then focus on the compound with the largest

subscript for the element in question to see if you can use that

subscript as the coefficient for the other compound.

3. If free, uncombined elements appear on either side, balance them

last.


As it now stands, the “equation” does not satisfy the Law of Conservation of Matter because there are two H atoms in the H2 molecule and three Cl atoms in one formula unit of AlCl3 (product side), but only one H atom and one Cl atom in the HCl molecule (reactant side). Let’s first balance chlorine by putting a coefficient of 3 in front of HCl.

Now there are 3H on the left and 2H on the right. We can add hydrogen only as two atoms at a time (as H2 ) on the product side. So we need to find the least common multiple of 3 and 2, which is 6, in order to balance the H atoms. To get a total of 6H atoms on each side, we multiply the 3HCl by 2 and the lone H2 by 3. This now gives


C4H10 ! O2 88n 4CO2 ! 5H2O


5 " 2 # 10

4C, 10H, 13O

reactants products


C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O



2C4H10 ! 13 O2 88n 8CO2 ! 10H2O

2 [ C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O ]








reactants products





"2



10663_03_ch03_p081-114.indd 85 12/7/12 1:51 PM



C4H10 ! O2 88n 4CO2 ! 5H2O


5 " 2 # 10

4C, 10H, 13O

reactants products


C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O



2C4H10 ! 13 O2 88n 8CO2 ! 10H2O

2 [ C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O ]








reactants products





"2



10663_03_ch03_p081-114.indd 85 12/7/12 1:51 PM



C4H10 ! O2 88n 4CO2 ! 5H2O


5 " 2 # 10

4C, 10H, 13O

reactants products


C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O



2C4H10 ! 13 O2 88n 8CO2 ! 10H2O

2 [ C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O ]








reactants products





"2



10663_03_ch03_p081-114.indd 85 12/7/12 1:51 PM




Now Cl is again unbalanced (6Cl on the left, 3Cl on the right), but we can fix this by putting a coefficient of 2 in front of AlCl3 on the product side.

Now all elements except Al are balanced (1Al on the left, 2Al on the right); we complete the balancing by putting a coefficient of 2 in front of Al on the reactant side.


C4H10 ! O2 88n 4CO2 ! 5H2O


5 " 2 # 10

4C, 10H, 13O

reactants products


C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O



2C4H10 ! 13 O2 88n 8CO2 ! 10H2O

2 [ C4H10 ! 6.5 O2 88n 4CO2 ! 5H2O ]








reactants products





"2



10663_03_ch03_p081-114.indd 85 12/7/12 1:51 PM


Al ! 6HCl 88n 2AlCl3 ! 3H2


Al, 6H, 6Clatom count: 2Al, 6H, 6Cl

"2

2Al ! 6HCl 88n 2AlCl3 ! 3H2

2Al ! 6HCl 88n 2AlCl3 ! 3H2

Al ! 6HCl 88n 2AlCl3 ! 3H2

2Al, 6H, 6Clfinal atom count: 2Al, 6H, 6Cl

"2

Practice exercise• Balance the following chemical equations

1. P4 + Cl2 → PCl32. RbOH + SO2 → Rb2SO3 + H2O3. P4O10 + Ca(OH)2 → Ca3(PO4)2 + H2O4. Na + O2 → Na2O5. Mg3N2 + H2O → NH3 + Mg(OH)2 6. LiCl + Pb(NO3)2 → PbCl2 + LiNO3 7. H2O + KO2 → KOH + O2

8. H2SO4 + NH3 → (NH4)2SO4

9. Na + O2 → Na2O2

10. P4 +O2 → P4O10

11. Ca(HCO3)2 + Na2CO3 → CaCO3 + NaHCO3 12. NH3 +O2 → NO + H2O13. Rb + H2O → RbOH + H2 14. Fe2O3 + CO → Fe + CO2

15. (NH4)2Cr2O7 → N2 + H2O + Cr2O3 16. Al + Cr2O3 → Al2O3 + Cr


Calculations Based on Chemical EquationsLet’s use chemical equations to calculate the relative amounts of substances involved in chemical reactions. Consider the combustion of methane in oxygen. The balanced chemical equation for that reaction is

CH4 + 2O2 ⟶ CO2 + 2H2OHow many O2 molecules react with 47 CH4 molecules according to the preceding equation?The balanced equation tells us that one CH4 molecule reacts with two O2 molecules and yields one CO2 and two molecules of water. We can hence construct the proportion

1 mol CH4 2 mol O2

47 mol CH4 X mol O2

=

1 mol CH4 2 mol O2

47 mol CH4 =mol O2 x = 94 mol O2

Limiting ReactantsSuppose you wish to make several sandwiches using one slice of cheese and two slices of bread for each.

Using Bd = breadCh = cheeseBd2Ch = sandwich

the recipe for making a sandwich can be represented like a chemical equation:

2Bd + Ch ⟶ Bd2Ch

If you have 10 slices of bread and 7 slices of cheese, you can make only 5 sandwiches and will have 2 slices of cheese left over. The amount of bread available limits the number of sandwiches.

10Bd + 7Ch ⟶ 5Bd2Ch + 2Ch

The reactant that is completely consumed in a reaction is called the limiting reactant because it limits the amount of product formed. The other reactants are sometimes called excess reactants

A chemical reaction stops as soon as any reactant is consumed, leaving the excess reactants as leftovers.

+ →

Limiting ReactantsSuppose, for example, we have a mixture of 10 mol H2 and 7 mol O2, which react to form water:

2H2(g) + O2(g) ⟶ 2 H2O(g)

As 7 mol O2 are available before the reaction, 7 mol O2 - 5 mol O2 = 2 mol O2 are present when all the H2 is consumed.

In this case H2 is the limiting reactant while O2 is the excess reactant.

Because 2 mol H2 react with 1 mol O2, the number of moles of O2 needed to react with 10 mol H2 is

1 mol O2

2 mol H210 mol H2 = 5 mol O2( )Moles of O2 =100 CHAPTER 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Because 7 mol O2 is available at the start of the reaction,is present when all the H2 is consumed.

The reactant that is completely consumed in a reaction is called the limiting reactantbecause it determines, or limits, the amount of product formed. The other reactants are

sometimes called excess reactants. In our example,shown in ! FIGURE 3.17, H2 is the limiting reactant,which means that once all the H2 has been consumed,the reaction stops. The excess reactant is O2; some is leftover when the reaction stops.

There are no restrictions on the starting amountsof reactants in any reaction. Indeed, many reactions arecarried out using an excess of one reactant. The quanti-ties of reactants consumed and products formed, how-ever, are restricted by the quantity of the limitingreactant. For example, when a combustion reactiontakes place in the open air, oxygen is plentiful and is

therefore the excess reactant. If you run out of gasoline while driving, the car stops be-cause the gasoline is the limiting reactant in the combustion reaction that moves the car.

Before we leave the example illustrated in Figure 3.17, let’s summarize the data:

= 2 mol O2

7 mol O2 - 5 mol O2

10 H2O and 2 O2 (no H2 molecules)10 H2 and 7 O2

After reactionBefore reaction

" FIGURE 3.17 Limiting reactant.Because H2 is completely consumed, it isthe limiting reactant. Because some O2 isleft over after the reaction is complete, O2is the excess reactant. The amount of H2Oformed depends on the amount of limitingreactant, H2.

G O F I G U R EIf O2 had been the limitingreactant, how many moles of H2O would have formed?

2 H2(g) ! O2(g) ¡ 2 H2O(g)

Initial quantities: 10 mol 7 mol 0 molChange (reaction): -10 mol -5 mol +10 molFinal quantities: 0 mol 2 mol 10 mol

The second line in the table (Change) summarizes the amounts of reactants con-sumed (where this consumption is indicated by the minus signs) and the amount of theproduct formed (indicated by the plus sign). These quantities are restricted by the quan-tity of the limiting reactant and depend on the coefficients in the balanced equation.The mole ratio conforms to the ratio of the coefficients in thebalanced equation, 2:1:2. The final quantities, which depend on the initial quantitiesand their changes, are found by adding the initial quantity and change quantity for eachcolumn. None of the limiting reactant (H2) remains at the end of the reaction. What re-mains is 2 mol O2 (excess reactant) and 10 mol H2O (product).

H2:O2:H2O = 10:5:10

The number of moles of H2 needed for complete consumption of3.0 mol of N2 is: Moles H2 = (3.0 mol N2)¢ 3 mol H2

1 mol N2≤ = 9.0 mol H2

Because only 6.0 mol H2 is available, we will run out of H2 beforethe N2 is gone, which tells us that H2 is the limiting reactant.Therefore, we use the quantity of H2 to calculate the quantity ofNH3 produced:

Moles NH3 = (6.0 mol H2)¢ 2 mol NH3

3 mol H2≤ = 4.0 mol NH3

SAMPLE EXERCISE 3.18 Calculating the Amount of Product Formed from a Limiting Reactant

Analyze We are asked to calculate the number of moles of product,NH3, given the quantities of each reactant, N2 and H2, available in areaction. This is a limiting reactant problem.

Plan If we assume one reactant is completely consumed, we can cal-culate how much of the second reactant is needed. By comparing thecalculated quantity of the second reactant with the amount available,we can determine which reactant is limiting. We then proceed withthe calculation, using the quantity of the limiting reactant.

The most important commercial process for converting N2 from the air into nitrogen-containing com-pounds is based on the reaction of N2 and H2 to form ammonia (NH3):

How many moles of can be formed from 3.0 mol of N2 and 6.0 mol of H2?

SOLUTION

NH3

N2(g) + 3 H2(g) ¡ 2 NH3(g)

Solve

100 CHAPTER 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Because 7 mol O2 is available at the start of the reaction,is present when all the H2 is consumed.

The reactant that is completely consumed in a reaction is called the limiting reactantbecause it determines, or limits, the amount of product formed. The other reactants are

sometimes called excess reactants. In our example,shown in ! FIGURE 3.17, H2 is the limiting reactant,which means that once all the H2 has been consumed,the reaction stops. The excess reactant is O2; some is leftover when the reaction stops.

There are no restrictions on the starting amountsof reactants in any reaction. Indeed, many reactions arecarried out using an excess of one reactant. The quanti-ties of reactants consumed and products formed, how-ever, are restricted by the quantity of the limitingreactant. For example, when a combustion reactiontakes place in the open air, oxygen is plentiful and is

therefore the excess reactant. If you run out of gasoline while driving, the car stops be-cause the gasoline is the limiting reactant in the combustion reaction that moves the car.

Before we leave the example illustrated in Figure 3.17, let’s summarize the data:

= 2 mol O2

7 mol O2 - 5 mol O2

10 H2O and 2 O2 (no H2 molecules)10 H2 and 7 O2

After reactionBefore reaction

" FIGURE 3.17 Limiting reactant.Because H2 is completely consumed, it isthe limiting reactant. Because some O2 isleft over after the reaction is complete, O2is the excess reactant. The amount of H2Oformed depends on the amount of limitingreactant, H2.

G O F I G U R EIf O2 had been the limitingreactant, how many moles of H2O would have formed?

2 H2(g) ! O2(g) ¡ 2 H2O(g)

Initial quantities: 10 mol 7 mol 0 molChange (reaction): -10 mol -5 mol +10 molFinal quantities: 0 mol 2 mol 10 mol

The second line in the table (Change) summarizes the amounts of reactants con-sumed (where this consumption is indicated by the minus signs) and the amount of theproduct formed (indicated by the plus sign). These quantities are restricted by the quan-tity of the limiting reactant and depend on the coefficients in the balanced equation.The mole ratio conforms to the ratio of the coefficients in thebalanced equation, 2:1:2. The final quantities, which depend on the initial quantitiesand their changes, are found by adding the initial quantity and change quantity for eachcolumn. None of the limiting reactant (H2) remains at the end of the reaction. What re-mains is 2 mol O2 (excess reactant) and 10 mol H2O (product).

H2:O2:H2O = 10:5:10

The number of moles of H2 needed for complete consumption of3.0 mol of N2 is: Moles H2 = (3.0 mol N2)¢ 3 mol H2

1 mol N2≤ = 9.0 mol H2

Because only 6.0 mol H2 is available, we will run out of H2 beforethe N2 is gone, which tells us that H2 is the limiting reactant.Therefore, we use the quantity of H2 to calculate the quantity ofNH3 produced:

Moles NH3 = (6.0 mol H2)¢ 2 mol NH3

3 mol H2≤ = 4.0 mol NH3

SAMPLE EXERCISE 3.18 Calculating the Amount of Product Formed from a Limiting Reactant

Analyze We are asked to calculate the number of moles of product,NH3, given the quantities of each reactant, N2 and H2, available in areaction. This is a limiting reactant problem.

Plan If we assume one reactant is completely consumed, we can cal-culate how much of the second reactant is needed. By comparing thecalculated quantity of the second reactant with the amount available,we can determine which reactant is limiting. We then proceed withthe calculation, using the quantity of the limiting reactant.

The most important commercial process for converting N2 from the air into nitrogen-containing com-pounds is based on the reaction of N2 and H2 to form ammonia (NH3):

How many moles of can be formed from 3.0 mol of N2 and 6.0 mol of H2?

SOLUTION

NH3

N2(g) + 3 H2(g) ¡ 2 NH3(g)

Solve

Before reaction After reaction

Limiting ReactantsSummarizing the data from the example before

2H2(g) + O2(g) ⟶ 2 H2O(g)

Before the reaction 10 mol 7 mol 0 mol

Change (reaction): -10 mol -5 mol +10 mol

After the reaction: 0 mol 2 mol 10 mol

The second line in the table (reaction) summarizes the amounts of reactants consumed (minus signs) and the amount of the product formed (plus sign). These quantities are restricted by the quantity of the limiting reactant and depend on the coefficients in the balanced equation.

The mole ratio H2:O2:H2O = 10:5:10 conforms to the ratio of the coefficients in the balanced equation, 2:1:2.

limiting reactant excess reactant product

Sample exercise (Limiting Reactants)

• A process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of of N2 and H2 to form ammonia (NH3):

N2(g) + 3 H2(g) → 2 NH3(g) How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?

The number of moles of H2 needed for complete consumption of 3.0 mol of N is:

Because only 6.0 mol H2 are available, we will run out of H2 before the N2 is gone, which tells us that H2 is the limiting reactant. Therefore, we use the quantity of H2 to calculate the quantity NH3 produced:

3 mol H21 mol H2

3 mol N2 = 9 mol H2( )Moles of H2 =

2 mol NH33 mol H2

6 mol H2 = 4 mol NH3( )Moles of NH3 =

N2(g) + 3 H2(g) → 2 NH3(g)Before the reaction 3 mol 6 mol 0 molChange (reaction): - 2 mol - 6 mol + 4 mol

After the reaction: 1 mol 0 mol 4 mol

Summarizing the data

Sample exercise (Limiting Reactants)

• The reaction 2H2(g) + O2(g) → 2H2O(g) is used to produce electricity in a hydrogen fuel cell. Suppose a fuel cell contains 150 g of H2(g) and 1500 g of O2(g). How many grams of water can form?

From the balanced equation, we have the stoichiometric relations 2mol H2 : ︎ 1molO2 : ︎ 2molH2O Using the molar mass of each substance, we calculate the number of moles of each reactant:

The balanced equation indicates that the reaction requires 2 mol of H2 for every mol of O2. Therefore, for all the O2 to completely react, we would need 2 x 47 = 94 mol of H2. Since there are only 75 mol of H2, all of the O2 cannot react, so it is the excess reactant, and H2 is the limiting reactant.

We use the limiting reactant H2 to calculate the quantity of water formed, 75 mol H2O. Then we can calculate the grams of water formed:

Grams of H2O = 75 mol x 18.0 g/mol = 1400 g H2O

1 mol H2

2.00 g H2150 g H2 = 75 mol H2( )Moles of H2 =

1 mol O2

32.0 g O21500 g O2 = 47 mol O2( )Moles of O2 =

Before the reaction 75 mol 47 mol 0 molChange (reaction): - 75 mol - 37,5 mol + 75 mol

After the reaction: 0 mol 9,5 mol 75 mol

2H2(g) + O2(g) → 2H2O(g)

m

mol x Fw

Practice exercises

1. When1.50 mol of Al and 3.00 mol of Cl2 combine in the reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)

a. Which is the limiting reactant?b. How many moles of AlCl3 are formed? c. How many moles of the excess reactant remain at the end of the reaction?

Answers: (a) Al, (b) 1.50 mol, (c) 0.75 mol Cl2

2. 2.00 g of zinc metal are placed in an aqueous solution containing 2.50 g of silver nitrate, the reaction isZn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq)

a. Which is the limiting reactant?b. How many grams of Ag form? c. How many grams of Zn(NO3)2 form?

Answers: (a)AgNO3,(b)1.59g,(c)1.39g,(d)1.52gZn

3. Sodium hydroxide reacts with carbon dioxide as follows: 2 NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)

a. Which is the limiting reactant when 1.85 mol NaOH and 1.00 mol CO2 are allowed to react?b. How many moles of Na2CO3 can be produced? c. How many moles of the excess reactant remain after the completion of the reaction?

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06.2 - Chemical equations and reaction stoichiometry · Chemical Equations HH H O O O C O H H H H H C O O O reaction stoichiometry: the quantitative relationships among substances