Chapter9_Equivalents, Gram Equivalent Mass, Normality

© Fred H Watson 2003

©FHWatson 159

Chapter 9. Equivalents, Gram Equivalent Mass (GEM), Normality

Thus far, we have followed a fixed procedure in working stoichiometry problems which utilizes balanced chemical equations. The procedure is based on our adoption of the mole as the fundamental unit for expressing the quantities of matter in chemical reactions. In other words, we read our reactions in terms of moles. Thus the reaction, 2 HCl(aq) + Ba(OH)2(sol) 2 H2O(aq) + BaCl2(aq), may be read as follows:

“If two moles of aqueous hydrochloric acid reacts with one mole of solid barium hydroxide, then two moles of water and one mole of aqueous barium chloride should be produced,” or, “If two moles of HCl(aq) are consumed by this reaction, then one mole of Ba(OH)2(sol) will be consumed, two moles of water should be produced, and one mole of BaCl2(aq)should also be produced.”

In order to perform stoichiometric calculations, we need the balanced equation from which to extract the appropriate mole ratios of the reagents and products in the reaction. In these notes, calculations based on balanced chemical equations will be referred to as the Mole Method. In many cases, the equations are easy to balance with respect to mass and charge, and we do so by inspection (also called trial-and error). When we encountered aqueous redox reactions, we came across a class of reactions that were more difficult to balance in general. We thus devised a ten-step procedure for charge and mass balancing redox reactions. In early chemistry texts, a method was introduced that removed the necessity of balancing chemical equations in order to perform stoichiometry calculations. This method requires that one properly define the mass in grams of an equivalent (the gram equivalent mass, GEM) of each participant (reactant or product) in the reaction involved in the calculation. This method will be referred to, henceforth, as the Equivalent Method. It should be emphasized that for any substance, a mole of it is defined the same regardless of whatever reaction it is participating in. A mole is a mole is a mole. However, an equivalent of any substance is defined with regard to the particular reaction in which it is participating. Thus the definition of an equivalent of some chemical species is not a general definition but rather depends on the particular reaction being studied. An equivalent is defined as follows.

equivalent ≡ the amount of a species which is chemically equivalent to one mole of electrons or to one mole of H+

(aq) ions or to half a mole of H2(gas). “Chemically equivalent” means “will react with” or “will produce.”

Specific definitions depend on whether the reaction being studied is a non-redox reaction or a redox reaction. For non-redox reactions that go essentially to completion (such as acid/base reactions, precipitation reactions, and reactions that produce an escaping gas), the GEM of a salt is defined as follows.

GEMsalt ≡ charge) positive (total

GMM = charge negative total

GMM


©FHWatson 160

Note: For a neutral salt, the total positive charge equals the absolute value of the total negative charge. For example, in Fe2(SO4)3, the total positive charge = 2(+3) = 6, and the absolute value of the total negative charge = │3(–2) │= 6. Therefore, in a non-redox reaction, the GEM of iron(III)sulfate equals the GMM/6 = 55.847/6 = 9.308 grams/equivalent = 9.308 g/eq. In keeping with this definition of a salt, we can define the GEMs of acids and bases as follows.

GEMacid ≡ protons ionized ofnumber

GMM ; GEMbase ≡ hydroxides ionized ofnumber

GMM .

• Rxn (1) H3PO4(aq) + NaOH NaH2PO4(aq) + H2O(aq), GEMphosphoric acid = GMM/(3–2) = GMM/1. • Rxn (2) H3PO4(aq) + NaOH Na2HPO4(aq) + H2O(aq), GEMphosphoric acid = GMM/(3–1) = GMM/2. • Rxn (3) H3PO4(aq) + NaOH Na3PO4(aq) + H2O(aq), GEMphosphoric acid = GMM/(3–0) = GMM/3. Suppose that we have a 0.25 M aqueous solution of phosphoric acid. This is its molarity no matter what reaction is involved since molarity is defined as the number of moles of solute per liter of solution. The normality (N) of a solution is defined as the number of equivalents of solute per liter of solution.

normality (N) ≡ the number of equivalents of solute/ the number of liters of solution.

• In Rxn (1), GEM = GMM; 1 mole = 1 eq; N = 1 × M = mole

eq 1 × Lmoles 0.25 = 0.25 eq/L = 0.25 N.

• In Rxn (2) above, GEM = GMM/2; 1 mole = 2 eqs; N = 2 × M = 2 × 0.25 = 0.50 eq/L = 0.50 N. • In Rxn (3) above, GEM = GMM/3; 1 mole = 3 eqs; N = 3 × M = 3 × 0.25 = 0.75 eq/L = 0.75 N. Note that in the three acid/base reactions given, the GEM for sodium hydroxide equals its GMM since it is a monohydroxy base and can utilize only one hydroxide ion. Thus, in each of these reactions, 1 eq = 1 mole of NaOH. If in aqueous solution, its normality will equal its molarity. The three phosphoric acid/sodium hydroxide reactions were deliberately not balanced, because the idea of defining equivalents is to avoid having to balance the reaction equation. Thus one may say that, “When one equivalent of a given reactant is consumed, one equivalent of each of the other reactants will be consumed, and one equivalent of each product should be produced. Whereas we always label moles in order to identify the chemical species, we usually do not label equivalents since the number of equivalents applies to every participating species in the chemical reaction. Thus, in Rxn (1), 1 molephosphoric acid = 1 equivalent; in Rxn (2), 1 molephosphoric acid = 2 eq; and in Rxn (3), 1 molephosphoric acid = 3 eq. The GEM of molecular compounds is more difficult to define. For the very important molecular compound and very weak electrolyte, water, its GEM in acid/base reactions is defined as its GMM/1 in order to account for the fact that it weakly ionizes into H+

(aq) and OH–(aq) ions.


©FHWatson 161

For species participating in redox reactions (reactants or products), the GEM is defined as follows.

GEMredox = numberoxidation in change total

GMM =

atom)per number oxidation in (changeformula) in the appearselement central the timesof(number GMM

×

Consider the same 0.25 M phosphoric acid solution as before, but engaged in the following half-reactions. We can tell that these half-reactions are not part of acid base reactions because the oxidation numbers for phosphorus are changing.

(4) (aq)4

)2(–5)(

3

)1(OPH

++

3(aq)

)2(–3)(

3

)1(OPH

++

; GEMphosphoric acid = GMM/1(5–3) = GMM/2, 1 mole = 2 eqs, N = 2 × M = 2 × 0.25 = 0.50 eq/L = 0.50 N.

(5) (aq)4

)2(–5)(

3

)1(OPH

++

3(gas)

)1(–3)(FP

+

; GEMphosphoric acid = GMM/1(5–3) = GMM/2, 1 mole = 2 eqs, N = 2 × M = 2 × 0.25 = 0.50 eq/L = 0.50 N.

(6) (aq)4

)2(–5)(

3

)1(OPH

++

4(sol)

)0(P ;

GEMphosphoric acid = GMM/1(5–0) = GMM/5, 1 mole = 5 eqs, N = 5 × M = 5 × 0.25 = 1.25 eq/L = 1.25 N. Likewise, we may define GEMs of the species produced by these half-reactions. GEMphosphorous acid = GMM/1(5–3) = GMM/2, 1 mole = 2 eqs, N = 2 × M. GEMphosphorus trifluoride = GMM/1(5–3) = GMM/2, 1 mole = 2 eqs, N = 2 × M. GEMsolid phosphorus = GMM/4(5–0) = GMM/20, 1 mole = 20 eqs, N = 20 × M. The steps for doing stoichiometric calculation by the Equivalent Method based on an unbalanced chemical reaction are always the same, as follows. Step #1: Determine the GEM of all species mentioned in the problem statement. If the information about a given species is given or requested in grams, calculate the numerical value of its GEM. If the amount of a given substance is given or requested in moles, simply determine how many equivalents are equal to a mole of it. If the molarity of a species in solution is given or requested, determine its normality. Step #2: Write the starting information. Step #3: Convert to equivalents. Step #4: Convert to the requested amount of the requested species in the requested units. These steps will be demonstrated by solving by the Equivalent Method the same problems that were solved by the Mole Method in the hand-out entitled Chemical Reactions in Aqueous Solutions.


©FHWatson 162

Non-Redox Problem #1 Consider the reaction of aqueous acetic acid with solid barium hydroxide. If one reacts 600 mL of 0.800 M acetic acid solution with 50.0 grams of solid barium hydroxide, how many molecules of water and how many grams of barium acetate will be produced? Step #1: Determine the GEM of all species mentioned in the problem statement. If the information about a given species is given or requested in grams, calculate the numerical value of its GEM. If the amount of a given substance is given or requested in moles, simply determine how many equivalents are equal to a mole of it. If the molarity of a species in solution is given or requested, determine its normality.

HC2H3O2(aq) + Ba(OH)2(sol) H2O(aq) + Ba(C2H3O2)2(aq). (unbalanced)

GEMacetic acid = ⎟⎠⎞

⎜⎝⎛

1GMM ; 1 mole = 1eq: N = 1 × M = 0.800 N.

GEMbarium hydroxide = ⎟⎠⎞

⎜⎝⎛

2GMM = ⎟

⎠⎞

⎜⎝⎛

2Ba(OH) g 171.3 2 = 85.67 g/eq.

GEMwater = ⎟⎠⎞

⎜⎝⎛

1GMM ; 1 mole = 1 eq.

GEMbarium acetate = ⎟⎟⎠

⎞⎜⎜⎝

⎛2(1)

GMM = ⎟⎠⎞

⎜⎝⎛

2)OHBa(C g 255.4 2232 = 127.7 g/eq.

Determination of the Limiting Reagent Step#2 Step#3 Step#4

(600 mL) ⎟⎟⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎠⎞

⎜⎝⎛

Leqs 0.800

⎟⎟⎠

⎞⎜⎜⎝

⎛eqBa(OH) g 85.67 2 = 41.1 g Ba(OH)2 needed. Since we have 50.0

grams of barium hydroxide, it is the excess reagent and acetic acid is limiting. Solution to Non-Redox Problem #1 Step#2 Step#3 Step#4

(600 mL) ⎟⎟⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎠⎞

⎜⎝⎛

Leqs 0.800

⎟⎟⎠

⎞⎜⎜⎝

⎛eq 1

OH mole 1 2⎟⎟⎠

⎞⎜⎜⎝

⎛ ×mole 1

molecules 10 6.022 23

= 2.89 × 1023 molecules.

Step#2 Step#3 Step#4

(600 mL) ⎟⎟⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎠⎞

⎜⎝⎛

Leqs 0.800

⎟⎟⎠

⎞⎜⎜⎝

⎛eq

)OHBa(C g 127.7 2232 = 61.3 g Ba(C2H3O2)2.


©FHWatson 163

Non-Redox Problem #2 If one reacts 500 mL of 0.250 M aqueous sodium hydroxide solution with an aqueous solution that contains 7.13 g of aluminum sulfate, how many grams of aluminum hydroxide solid will form and how many moles of sodium sulfate will be produced? Step #1: Determine the GEM of all species mentioned in the problem statement. If the information about a given species is given or requested in grams, calculate the numerical value of its GEM. If the amount of a given substance is given or requested in moles, simply determine how many equivalents are equal to a mole of it. If the molarity of a species in solution is given or requested, determine its normality.

Al2(SO4)3(aq) + NaOH(aq) Al(OH)3(sol) ↓ + Na2(SO4)(aq). (unbalanced)

GEMaluminum sulfate = ⎟⎟⎠

⎞⎜⎜⎝

⎛2(3)

GMM = ⎟⎠⎞

⎜⎝⎛

6342.1 = 57.0 g/eq.

GEMsodium hydroxide = ⎟⎟⎠

⎞⎜⎜⎝

⎛1(1)

GMM ; 1molesodium hydroxide = 1 eq; N = 1 × 0.250 = 0.250 N.

GEMaluminum hydroxide = ⎟⎟⎠

⎞⎜⎜⎝

⎛1(3)

GMM = ⎟⎠⎞

⎜⎝⎛

378.0 = 26.0 g/eq.

GEMsodium sulfate = ⎟⎟⎠

⎞⎜⎜⎝

⎛2(1)

GMM ; 1molesodium sulfate = 2 eqs.

Determination of the Limiting Reagent Step#2 Step#3 Step#4

(7.13 g Al2(SO4)3) ⎟⎟⎠

⎞⎜⎜⎝

⎛

342 )(SOAl g 57.0eq 1

⎟⎟⎠

⎞⎜⎜⎝

⎛eq 0.250

NaOH L 1⎟⎠⎞

⎜⎝⎛

L 10mL 1

3– = 500 mL of the NaOH solution needed.

Since we have 500 mL of the NaOH solution, neither reagent is in excess. Both are limiting. Solution to Non-Redox Problem #2 Step#2 Step#3 Step#4

(7.13 g Al2(SO4)3) ⎟⎟⎠

⎞⎜⎜⎝

⎛

342 )(SOAl g 57.0eq 1

⎟⎟⎠

⎞⎜⎜⎝

⎛eq 1Al(OH) g 26.0 3 = 3.25 g Al(OH)3.


(7.13 g Al2(SO4)3) ⎟⎟⎠

⎞⎜⎜⎝

⎛

342 )(SOAl g 57.0eq 1

⎟⎟⎠

⎞⎜⎜⎝

⎛eq 2

)(SONa mole 1 42 = 0.0625 moles Na2(SO4).

Alternate Solution to Non-Redox Problem #2 Step#2 Step#3 Step#4

(500 mL NaOH) ⎟⎟⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎠⎞

⎜⎝⎛

NaOH Leqs 0.250

⎟⎟⎠

⎞⎜⎜⎝

⎛eq 1Al(OH) g 26.0 3 = 3.25 g Al(OH)3.


(500 mL NaOH) ⎟⎟⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎠⎞

⎜⎝⎛

NaOH Leqs 0.250 ⎟⎟

⎠

⎞⎜⎜⎝

⎛eq 2

)(SONa mole 1 42 = 0.0625 moles Na2(SO4).


©FHWatson 164

Non-Redox Problem #3 If one titrates 100 mL of 0.540 M H3PO4(aq) against 0.333 M Ca(OH)2(aq), how many mL of the basic solution will be required to reach the ultimate end-point. Step #1: Determine the GEM of all species mentioned in the problem statement. If the information about a given species is given or requested in grams, calculate the numerical value of its GEM. If the amount of a given substance is given or requested in moles, simply determine how many equivalents are equal to a mole of it. If the molarity of a species in solution is given or requested, determine its normality.

H3PO4(aq) + Ca(OH)2(aq) H2O(aq) + Ca3(PO4)2(aq). (unbalanced)

GEMphosphoric acid = ⎟⎠⎞

⎜⎝⎛

3GMM ; 1 mole = 3 eqs.; N = 3 × M = 3 × 0.540 = 1.62 N.

GEMcalcium hydroxide = ⎟⎠⎞

⎜⎝⎛

2GMM ; 1 mole = 2 eqs.; N = 2 × M = 2 × 0.333 = 0.666 N.

Note: Since water and calcium phosphate are not mentioned in the problem statement, we do not have to define their GEMs. Solution to Non-Redox Problem #2 Step#2 Step#3 Step#4

(100 mL acid) ⎟⎟⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎠⎞

⎜⎝⎛

acid Leqs 1.62

⎟⎟⎠

⎞⎜⎜⎝

⎛eqs 0.666

base L⎟⎠⎞

⎜⎝⎛

L 10mL 1

3– = 243 mL base.

Alternate Solution to Non-Redox problem #3. Since the units for normality are eqs/L, then the volume of a sample (in liters) times its normality equals the number of equivalents in the sample. In an acid/base reaction, we know that one equivalent of acid will react with one equivalent of base. Thus, we may write the formula: VAMA = VBMB. Using this equation,

VB = ⎟⎟⎠

⎞⎜⎜⎝

⎛

A

B

NN VA = ⎟

⎠⎞

⎜⎝⎛

N 0.666N 62.1 (100 mL) = 243 mL.


©FHWatson 165

Redox Problem #1 Consider the unbalanced reaction,

(aq)

1)(OClK

+

+ 4(aq)

7)(OMnK

+

⎯⎯→⎯basic3(aq)

5)(OClK

+

+ 2(sol)

4)(OMn

+

. (unbalanced) If one reacts 500 mL of 0.200 M aqueous KClO with 700 mL of 0.175 M aqueous KMnO4, how many moles of potassium chlorate and how many grams of MnO2 should be produced. Step #1: Determine the GEM of all species mentioned in the problem statement. If the information about a given species is given or requested in grams, calculate the numerical value of its GEM. If the amount of a given substance is given or requested in moles, simply determine how many equivalents are equal to a mole of it. If the molarity of a species in solution is given or requested, determine its normality.

GEMpotassium hypochlorite = ⎟⎟⎠

⎞⎜⎜⎝

⎛1) - 1(5

GMM⎟⎠⎞

⎜⎝⎛

4GMM ; 1 mole = 4 eqs; N = 4 × 0.200 = 0.800 N.

GEMpotassium pemanganate = ⎟⎟⎠

⎞⎜⎜⎝

⎛4) - 1(7

GMM⎟⎠⎞

⎜⎝⎛

3GMM ; 1 mole = 3 eqs; N = 3 × 0.175 = 0.525 N.

GEMpotassium chlorate = ⎟⎟⎠

⎞⎜⎜⎝

⎛1) - 1(5

GMM⎟⎠⎞

⎜⎝⎛

4GMM ; 1 mole = 4 eqs.

GEMmanganese(IV) oxide = ⎟⎟⎠

⎞⎜⎜⎝

⎛4) - 1(7

GMM⎟⎠⎞

⎜⎝⎛

3GMM = ⎟⎟

⎠

⎞⎜⎜⎝

⎛3eqs

MnO g 86.94 2 = 29.0 g/eq.

Determining the Limiting Reagent Step#2 Step#3 Step#4

(500 mL KClO) ⎟⎟⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎠⎞

⎜⎝⎛

KClO Leqs 0.800

⎟⎟⎠

⎞⎜⎜⎝

⎛eq 525.0

KMnO L 1 4 ⎟⎠⎞

⎜⎝⎛

L 101mL

3- = 762 mL KMnO4 solution needed.

Since we have only 700 mL of the KMnO4 solution, KMnO4 is the limiting reagent. Solution to Redox Problem #3 Step#2 Step#3 Step#4

(700 mL KMnO4) ⎟⎟⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎟⎠

⎞⎜⎜⎝

⎛

4KMnO Leqs 0.525

⎟⎟⎠

⎞⎜⎜⎝

⎛eqs 4KClO mole 1 3 = 0.0919 moles KClO3.


(700 mL KMnO4) ⎟⎟⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎟⎠

⎞⎜⎜⎝

⎛

4KMnO Leqs 0.525

⎟⎟⎠

⎞⎜⎜⎝

⎛eqMnO g 29.0 2 = 10.7 g MnO2.


©FHWatson 166

Redox Problem #2 Consider the unbalanced reaction,

3(aq)3

3)()(NOCr

+

+ )(2

)1(

2 ONa aq

−

⎯⎯→⎯acid 7(aq)2

)6(

2 OCrNa+

+ (aq)

)2(

2 OH−

. (unbalanced) If one reacts 50.0 grams of Cr(NO3)3 with 0.400 M aqueous Na2O2 solution, how many mL of the sodium peroxide solution should be required, and how many grams of Na2Cr2O7 should be produced? Step #1: Determine the GEM of all species mentioned in the problem statement. If the information about a given species is given or requested in grams, calculate the numerical value of its GEM. If the amount of a given substance is given or requested in moles, simply determine how many equivalents are equal to a mole of it. If the molarity of a species in solution is given or requested, determine its normality. Since we have full information only on Cr(NO3)3, it is the limiting reagent.

GEMchromium(III)nitrate = ⎟⎟⎠

⎞⎜⎜⎝

⎛3) - 1(6

GMM = ⎟⎠⎞

⎜⎝⎛

3)Cr(NO g 238.0 33 = 79.33 g/eq.

GEMsodium peroxide = ⎟⎟⎠

⎞⎜⎜⎝

⎛(-2))- 2(-1

GMM = ⎟⎠⎞

⎜⎝⎛

2GMM ; 1 mole = 2 eqs; N = 2 × 0.400 = 0.800 N.

GEMsodium dichromate = ⎟⎟⎠

⎞⎜⎜⎝

⎛3) - 2(6

GMM = ⎟⎠⎞

⎜⎝⎛

6)O(CrNa g 262.0 722 = 43.67 g Na2(Cr2O7).

Solution to Redox Problem #3 Step#2 Step#3 Step#4

(50.0 g Cr(NO3)3) ⎟⎟⎠

⎞⎜⎜⎝

⎛

33 )Cr(NO g 79.33eq 1

⎟⎟⎠

⎞⎜⎜⎝

⎛eq 0.800ONa L 1 22 ⎟

⎠⎞

⎜⎝⎛

L 10mL 1

3- = 788 mL Na2O2(aq).


(50.0 g Cr(NO3)3) ⎟⎟⎠

⎞⎜⎜⎝

⎛

33 )Cr(NO g 79.33eq 1

⎟⎟⎠

⎞⎜⎜⎝

⎛eq

)O(CrNa g 43.67 722 = 27.5 g Na2(Cr2O7).


©FHWatson 167

Redox Problem #3 Consider the reaction,

3(aq)

5)(OClNa

+

⎯⎯→⎯basic4(aq)

7)(OClNa

+

+ )(2

)0(Cl gas . (unbalanced)

If 250 mL of 0.350 M sodium chlorate are consumed by this reaction, how many grams of sodium perchlorate and how many liters of chlorine gas at 25.0ºC and 1.00 atm should be produced? Given: dchlorine at 25.0ºC and 1.00 atm = 2.90 g/L. This problem was worked by the Mole Method as Redox Problem #3 in the document entitled Chemical Reactions in Aqueous Solution. It is unusual because the same substance is being oxidized and reduced. This is referred to as disproportionation. It presents difficulty in determining the GEMs. It also necessitates that we pre-determine how the material is proportioned to produce the two products.

For each oxidation, ClO3– d ClO4

–, the oxidation number increases by 7 – 5 = 2, thus 2 electrons are lost. For the reduction half-reaction, 2 ClO3

– d Cl2, the oxidation number decreases by 2(5 – 0) = 10, thus 10 electrons are gained. Therefore, to maintain neutrality, the reaction must produce 10/2 = 5 ClO4

–

ions for each Cl2 molecule produced. We thus determine in advance that for every 5 ClO3– ions that

produce ClO4– ions, 2 ClO3

– ions produce Cl2. Thus, 5/7 of the NaClO3 consumed produces NaClO4, and 2/7 of the NaClO3 produces Cl2. Obviously, sodium chlorate, being the only reagent, is the limiting reagent.

Step #1: Determine the GEM of all species mentioned in the problem statement. If the information about a given species is given or requested in grams, calculate the numerical value of its GEM. If the amount of a given substance is given or requested in moles, simply determine how many equivalents are equal to a mole of it. If the molarity of a species in solution is given or requested, determine its normality.

For production of NaClO4, GEMsodium chlorate = ⎟⎟⎠

⎞⎜⎜⎝

⎛ 5) - 1(7

GMM = ⎟⎠⎞

⎜⎝⎛

2GMM ; N = 2 × 0.350 = 0.700 N.

For production of Cl2, GEMsodium chlorate = ⎟⎟⎠

⎞⎜⎜⎝

⎛ 0) - 1(5

GMM = ⎟⎠⎞

⎜⎝⎛

5GMM ; N = 5 × 0.350 = 1.75 N.

GEMsodium perchlorate = ⎟⎟⎠

⎞⎜⎜⎝

⎛ 5) - 1(7

GMM = ⎟⎠⎞

⎜⎝⎛

2GMM = ⎟

⎠⎞

⎜⎝⎛

2NaClO g 112.5 4 = 56.25 g/eq.

GEMchlorine = ⎟⎟⎠

⎞⎜⎜⎝

⎛ 0) - 2(5

GMM = ⎟⎠⎞

⎜⎝⎛

10Cl g 70.91 2 = 7.091 g/eq.

Solution to Redox Problem #3 Step#2 Step#3 Step#4

⎟⎠⎞

⎜⎝⎛

75 (250 mL) ⎟⎟

⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎠⎞

⎜⎝⎛

Leqs 0.700

⎟⎟⎠

⎞⎜⎜⎝

⎛eqNaClO g 56.25 4 = 7.03 g NaClO4.


⎟⎠⎞

⎜⎝⎛

72 (250 mL) ⎟⎟

⎠

⎞⎜⎜⎝

⎛mL 1

L 10–3

⎟⎠⎞

⎜⎝⎛

Leqs 1.75

⎟⎟⎠

⎞⎜⎜⎝

⎛eq

Cl g 7.091 2⎟⎟⎠

⎞⎜⎜⎝

⎛

2

2

Cl g 2.90Cl L 1 = 0.306 L Cl2.

For redox reactions such as this, that exhibit disproportionation because the same substance is being oxidized and reduced, it is recommended that stoichiometry problems be worked by the Mole Method rather than by the Equivalent Method.

Documents

Chapter9_Equivalents, Gram Equivalent Mass, Normality