Stoichiometry of Formulas and Equations ปริมาณสัมพันธ์ของสูตรและสมการ

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3-1

Stoichiometry of Formulas and Equationsปริ�มาณสั�มพั�นธ์�ของสั�ตริและสัมการิ

Chapter 3


3-2

Mole - Mass Relationships in Chemical Systems

3.5 Fundamentals of Solution Stoichiometry

3.1 The Mole

3.2 Determining the Formula of an Unknown Compound

3.3 Writing and Balancing Chemical Equations

3.4 Calculating the Amounts of Reactant and Product


3-3

mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12.

This amount is 6.022x1023. The number is called Avogadro’s number and is abbreviated as N.

One mole (1 mol) contains 6.022x1023 entities (to four significant figures)


3-4

12 red marbles @ 7g each = 84g12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 1023 atoms Fe

32.07g S = 6.022 x 1023 atoms S

Figure 3.1 Counting objects of fixed relative mass.


3-5

Water 18.02 gCaCO3

100.09 g

Oxygen 32.00 g

Copper 63.55 g

One mole of common substances.

Figure 3.2


3-6

Table 3.1 Summary of Mass Terminology

Term Definition Unit

Isotopic mass Mass of an isotope of an element

amu

Atomic mass

Molecular (or formula) mass (also called molecular weight)

Molar mass (M)

(also called atomic weight)

(also called gram-molecular weight)

amu

amu

g/mol

Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance

Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit)

Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units)


3-7

Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol)

Oxygen (O)

Mass/mole of compound

6 atoms

96.00 g

Table 3.2Carbon (C) Hydrogen (H)

Atoms/moleculeof compound

Moles of atoms/mole of compound

Atoms/mole ofcompound

Mass/moleculeof compound

6 atoms 12 atoms

6 moles of atoms

12 moles of atoms

6 moles of atoms

6(6.022 x 1023) atoms

12(6.022 x 1023) atoms

6(6.022 x 1023) atoms

6(12.01 amu) =72.06 amu

12(1.008 amu) =12.10 amu

6(16.00 amu) =96.00 amu

72.06 g 12.10 g


3-8

Interconverting Moles, Mass, and Number of Chemical Entities

g

M


3-9

MASS(g)of elementMASS(g)

of element

AMOUNT(mol)of element

AMOUNT(mol)of element

ATOMSof element

ATOMSof element

Summary of the mass-mole-number relationships for elements.

M (g/mol)

Avogadro’s number(atoms/mol)

Figure 3.3


3-10

Sample Problem 3.1 Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element

PROBLEM:

PLAN:

SOLUTION:

amount(mol) of Ag

mass(g) of Ag

(a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342 mol of Ag?

(b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8 g of Fe?

(a) To convert mol of Ag to g we have to use the #g Ag/mol Ag, the molar mass M.

(b) To convert g of Fe to atoms we first have to find the #mols of Fe and then convert mols to atoms.

multiply by M of Ag (107.9g/mol)

0.0342 mol Ag xmol Ag

107.9 g Ag = 3.69 g Ag

PLAN: mass(g) of Fe

amount(mol) of Fe

atoms of Fe

SOLUTION: 95.8g Fe x55.85g Fe

mol Fe= 1.72 mol Fe

1.72mol Fe x 6.022x1023atoms Fe

mol Fe

= 1.04x1024 atoms Fe

divide by M of Fe (55.85g/mol)

multiply by 6.022x1023 atoms/mol


3-11

MASS(g)of compound

MASS(g)of compound

AMOUNT(mol)of compound

AMOUNT(mol)of compound

MOLECULES(or formula units)

of compound


of compound

AMOUNT(mol)of elements in

compound

AMOUNT(mol)of elements in

compound

Avogadro’s number(molecules/mol)

M (g/mol) chemical

formula

Summary of the mass-mole-number relationships for compounds.

Figure 3.3


3-12

Sample Problem 3.2 Calculating the Moles and Number of Formula Units in a Given Mass of a Compound

PROBLEM:

PLAN:

SOLUTION:

Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, first extinguishers, and smelling salts. How many formula unit are in 41.6 g of ammonium carbonate?

mass(g) of (NH4)2CO3

number of (NH4)2CO3 formula units

amount(mol) of (NH4)2CO3

After writing the formula for the compound, we find its M by adding the masses of the elements. Convert the given mass, 41.6 g to mols using M and then the mols to formula units with Avogadro’s number.

The formula is (NH4)2CO3.

M = (2 x 14.01 g/mol N)+(8 x 1.008 g/mol H) +(12.01 g/mol C)+(3 x 16.00 g/mol O)= 96.09 g/mol

41.6 g (NH4)2CO3 x

2.61x1023 formula units (NH4)2CO3

divide by M

multiply by 6.022x1023 formula units/mol

mol (NH4)2CO3

96.09 g (NH4)2CO3

6.022x1023 formula units (NH4)2CO3

mol (NH4)2CO3

x

=


3-13

Mass % of element X =

atoms of X in formula x atomic mass of X (amu)

molecular (or formula) mass of compound (amu)

x 100

Mass % of element X =

moles of X in formula x molar mass of X (amu)

molecular (or formula) mass of compound (amu)

x 100

Mass percent from the chemical formula


3-14

Sample Problem 3.3 Calculating the Mass Percents and Masses of Elements in a Sample of Compound

PLAN:

SOLUTION:

amount(mol) of element X in 1mol compound

mass(g) of X in 1mol of compound

mass % X in compound

PROBLEM: Glucose (C6H12O6) is the most important nutrient in the

living cell for generating chemical potential energy.

(a) What is the mass percent of each element in glucose?

(b) How many grams of carbon are in 16.55g of glucose?

mass fraction of X

multiply by M (g/mol) of X

divide by mass(g) of 1mol of compound

multiply by 100

We have to find the total mass of glucose and the masses of the constituent elements in order to relate them.

Per mole glucose there are 6 moles of C12 moles H6 moles O

(a)


3-15

Sample Problem 3.3 Calculating the Mass Percents and Masses of Elements in a Sample of Compound

continued

6 mol C x 12.01 g C

mol C = 72.06 g C

12 mol H x 1.008 g H

mol H = 12.096 g H

6 mol O x 16.00 g O

mol O= 96.00 g O M = 180.16 g/mol

(b)mass percent of C =

72.06 g C

180.16 g glucose= 0.3999 x 100 = 39.99 mass %C

mass percent of H =12.096 g H

180.16 g glucose= 0.06714 x 100 = 6.714 mass %H

mass percent of O =96.00 g O

180.16 g glucose= 0.5329 x 100 = 53.29 mass %O


3-16

Empirical and Molecular Formulas

Empirical Formula -

Molecular Formula -

The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms.

The formula of the compound as it exists, it may be a multiple of the empirical formula.


3-17

mass(g) of each element

Sample Problem 3.4 Determining the Empirical Formula from Masses of Elements

PROBLEM:

PLAN:

SOLUTION:

amount(mol) of each element

empirical formula

Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound?

preliminary formula

change to integer subscripts

use # of moles as subscripts

divide by M(g/mol)

Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula).

2.82 g Na xmol Na

22.99 g Na = 0.123 mol Na

4.35 g Cl x mol Cl

35.45 g Cl = 0.123 mol Cl

7.83 g O xmol O

16.00 g O= 0.489 mol O

Na1 Cl1 O3.98Na1 Cl1 O3.98NaClO4

NaClO4 is sodium

perchlorate.


3-18

assume 100g lactic acid and find the mass of each element

Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass

PROBLEM:

PLAN:

amount(mol) of each element

During physical activity. lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O.

(a) Determine the empirical formula of lactic acid.

(b) Determine the molecular formula.

preliminary formula

empirical formula

divide each mass by mol mass(M)

molecular formulause # mols as subscripts

convert to integer subscriptsdivide mol mass by mass of empirical formula to get a

multiplier


3-19

Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass

continued

SOLUTION: Assuming there are 100 g of lactic acid, the constituents are

40.0 g C x 6.71 g H x 53.3 g O xmol C

12.01g C

mol H

1.008 g H

mol O

16.00 g O

= 3.33 mol C = 6.66 mol H = 3.33 mol O

C3.33 H6.66 O3.33

3.33 3.33 3.33CH2O empirical

formula

mass of CH2O

molar mass of lactate 90.08 g

30.03 g3 C3H6O3 is the

molecular formula


3-20

Combustion train for the determination of the chemical composition of organic compounds.

Figure 3.4

CnHm + (n+ ) O2 = n CO(g) + H2O(g)m 2

m 2


3-21

difference (after-before) = mass of oxidized element

Sample Problem 3.6Determining a Molecular Formula from Combustion Analysis

PLAN:

find the mass of each element in its combustion product

molecular formula

PROBLEM: Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained:

mass of CO2 absorber after combustion = 85.35g

mass of CO2 absorber before combustion = 83.85g

mass of H2O absorber after combustion = 37.96g

mass of H2O absorber before combustion = 37.55g

What is the molecular formula of vitamin C?

find the molspreliminary

formulaempirical formula


3-22

SOLUTION:

CO285.35 g - 83.85 g = 1.50 g H2O

37.96 g - 37.55 g = 0.41 g

There are 12.01 g C per mol CO21.50 g CO 2 x

12.01 g CO2

44.01 g CO2

= 0.409 g C

0.41 g H2O x

2.016 g H2O

18.02 g H2O

= 0.046 g HThere are 2.016 g H per mol H2O

O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545

0.409 g C

12.01 g C

0.046 g H

1.008 g H

0.545 g O

16.00 g O= 0.0341 mol C = 0.0456 mol H = 0.0341 mol O

C1H1.3O1 C3H4O3

176.12 g/mol

88.06 g= 2.000

Molecular formular = C6H8O6

Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis

continued


3-23

Table 3.3 Some Compounds with Empirical Formula CH2O

(Composition by Mass: 40.0% C, 6.71% H, 53.3% O)

NameMolecular Formula

Whole-Number Multiple

M(g/mol) Use or Function

formaldehyde

acetic acid

lactic acid

erythrose

ribose

glucose

CH2O

C2H4O2

C3H6O3

C4H8O4

C5H10O5

C6H12O6

1

2

3

4

5

6

30.03

60.05

90.09

120.10

150.13

180.16

disinfectant; biological preservative

acetate polymers; vinegar (5%soln)

part of sugar metabolism

sour milk; forms in exercising muscle

component of nucleic acids and B2

major energy source of the cell

CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6


3-24

Table 3.4 Two Pairs of Constitutional Isomers

Property Ethanol Dimethyl Ether

M(g/mol)

Boiling Point

Density at 200C

Structural formulas

46.07

78.50C

0.789 g/mL(liquid)

46.07

-250C

0.00195 g/mL(gas)

Butane 2-Methylpropane

C4H10 C2H6O

58.12 58.12

Space-filling models

-0.50C

0.579 g/mL(gas)

-11.060C

0.549 g/mL(gas)

C C C C

H H

H

H H

H

H

H

H

H

C C C

H H

H

HC

H

H

H

H H

H

C C

H H

H

H H

OH C O

H

H

H

C

H

H

H


3-25

The formation of HF gas on the macroscopic and molecular levels.

Figure 3.6


3-26

A three-level view of the chemical reaction in a flashbulb.Figure 3.7


3-27

translate the statement

balance the atoms

specify states of matter

adjust the coefficients

check the atom balance


3-28

translate the statement

Sample Problem 3.7 Balancing Chemical Equations

PROBLEM:

PLAN:SOLUTION:

balance the atoms

specify states of matter

Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes

with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction.

adjust the coefficients

check the atom balance

C8H18 + O2 CO2 + H2O

C8H18 + O2 CO2 + H2O825/2 9

2C8H18 + 25O2 16CO2 + 18H2O

2C8H18 + 25O2 16CO2 + 18H2O

2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g)


3-29

MASS(g)of compound A

MASS(g)of compound A

AMOUNT(mol)of compound A

AMOUNT(mol)of compound A


of compound A


of compound A

Avogadro’s number

(molecules/mol)

M (g/mol) of compound A

molar ratio from

balanced equation

Summary of the mass-mole-number relationships in a chemical reaction.

Figure 3.8

MASS(g)of compound B

MASS(g)of compound B

AMOUNT(mol)of compound B

AMOUNT(mol)of compound B


of compound B


of compound B

Avogadro’s number

(molecules/mol)

M (g/mol) of compound B


3-30

Sample Problem 3.8 Calculating Amounts of Reactants and Products

PROBLEM: In a lifetime, the average American uses 1750 lb(794 g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistep process. After an initial grinding, the first step is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide.

(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?

PLAN: write and balance equation

find mols O2 find mols SO2

find g SO2

find mols Cu2O

find mols O2 find kg O2


3-31

SOLUTION:

Sample Problem 3.8 Calculating Amounts of Reactants and Products

continued

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

3mol O2

2mol Cu2S= 15.0mol O2

= 641g SO2

= 20.0mol Cu2O

= 0.959kg O2

kg O2

103g O2

20.0mol Cu2O x3mol O2

2mol Cu2O

32.00g O2

mol O2

10.0mol Cu2S x(a)

10.0mol Cu2S x2mol SO2

2mol Cu2S

64.07g SO2

mol SO2

(b)

2.86kg Cu2O x 103g Cu2O

kg Cu2O

mol Cu2O

143.10g Cu2O

(c)

(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?

(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?

(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?

x

x

xx


3-32

Sample Problem 3.9 Writing an Overall Equation for a Reaction Sequence

PROBLEM: Roasting is the first step in extracting copper from chalcocite, the ore used in the previous problem. In the next step, copper(I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step process.

PLAN:

write balanced equations for each step

cancel reactants and products common to both sides of the equations

sum the equations

SOLUTION:

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

2Cu2O(s) + 2C(s) 4Cu(s) + 2CO(g)

2Cu2S(s)+3O2(g)+2C(s) 4Cu(s)+2SO2(g)+2CO(g)


3-33

An ice cream sundae analogy for limiting reactions.Figure 3.10


3-34

Table 3.5 Information Contained in a Balanced Equation

Viewed in Terms of

ReactantsC3H8(g) + 5O2(g)

Products3CO2(g) + 4H2O(g)

molecules 3 molecules CO2 + 4 molecules H2O 1 molecule C3H8 + 5 molecules O2

amount (mol) 1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O

44.09 amu C3H8 + 160.00 amu O2 mass (amu) 132.03 amu CO2 + 72.06 amu H2O

mass (g) 44.09 g C3H8 + 160.00 g O2 132.03 g CO2 + 72.06 g H2O

total mass (g) 204.09 g 204.09 g


3-35

Sample Problem 3.10 Using Molecular Depictions to Solve a Limiting-Reactant Problem

PROBLEM: Nuclear engineers use chlorine trifluoride in the processing of uranium fuel for power plants. This extremely reactive substance is formed as a gas in special metal containers by the reaction of elemental chlorine and fluorine.

PLAN:

(a) Suppose the box shown at left represents a container of the reactant mixture before the reaction occurs (with chlorine colored green). Name the limiting reactant, and draw the container contents after the reaction is complete.

(b) When the reaction is run again with 0.750 mol of Cl2 and 3.00 mol of F2, what mass of chlorine trifluoride will be prepared?

Write a balanced chemical equation. Compare the number of molecules you have to the number needed for the products. Determine the reactant that is in excess. The other reactant is the limiting reactant.


3-36

Sample Problem 3.10 Using Molecular Depictions to Solve a Limiting-Reactant Problem

SOLUTION:

continued

Cl2(g) + 3F2(g) 2ClF3(g)

F2Cl2

(a) You need a ratio of 2 Cl and 6 F for the reaction. You have 6 Cl and 12 F. 6 Cl would require 18 F. 12 F need only 4 Cl (2 Cl2 molecules).

There isn’t enough F, therefore it must be the limiting reactant.

You will make 4 ClF2 molecules (4 Cl, 12 F) and have 2 Cl2 molecules left over.

(b) We know the molar ratio of F2/Cl2 should be 3/1.

3.00 mol F2

0.750 mol Cl2

=4

1

Since we find that the ratio is 4/1, that means F2 is in excess and Cl2 is the limiting reactant.

0.750 mol Cl2 x2 mol ClF3

1 mol Cl

92.5 g ClF3

1 mol ClF3

= 139 g ClF3x


3-37

Sample Problem 3.11 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant

PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N2H4) and dinitrogen

tetraoxide(N2O4), which ignite on contact to form nitrogen gas

and water vapor. How many grams of nitrogen gas form when 1.00x102 g of N2H4 and 2.00x102 g of N2O4 are mixed?

PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given.

In this case one of the reactants is in molar excess and the other will limit the extent of the reaction.

mol of N2 mol of N2

divide by M

molar ratio

mass of N2H4

mol of N2H4

mass of N2O4

mol of N2O4

limiting mol N2

g N2

multiply by M


3-38

Sample Problem 3.11 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant

continued

SOLUTION: N2H4(l) + N2O4(l) N2(g) + H2O(l)

1.00x102g N2H4 x = 3.12 mol N2H4

mol N2H4

32.05g N2H4

3.12mol N2H4 x = 4.68mol N2

3 mol N2

2mol N2H4

2.00x102g N2O4 x = 2.17mol N2O4

mol N2O4

92.02g N2O4

2.17mol N2O4 x = 6.51mol N2

3 mol N2

mol N2O4

N2H4 is the limiting reactant

because it produces less product, N2, than does N2O4.

4.68mol N2 xmol N2

28.02g N2 = 131g N2

2 43


3-39

A + B

(reactants)

C

(main product)

D

(side products)

The effect of side reactions on yield.Figure 3.11


3-40

Sample Problem 3.12 Calculating Percent Yield

PROBLEM: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0 kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process?

PLAN:

write balanced equation

find mol reactant & product

find g product predicted

percent yield

actual yield/theoretical yield x 100

SOLUTION:

SiO2(s) + 3C(s) SiC(s) + 2CO(g)

100.0 kg SiO2 x mol SiO2

60.09 g SiO2

103 g SiO2

kg SiO2 = 1664 mol SiO2

mol SiO2 = mol SiC = 1664

1664 mol SiC x40.10 g

SiCmol SiC

kg

103g= 66.73 kg

100 =77.0%51.4 kg

66.73 kg

x

x

x


3-41

Sample Problem 3.13 Calculating the Molarity of a Solution

PROBLEM: Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL?

mol of glycine

concentration(mol/mL) glycine

molarity(mol/L) glycine

SOLUTION:

PLAN: Molarity is the number of moles of solute per liter of solution.

0.715 mol glycine

495 mL soln

1000mL

1 L

divide by volume

103mL = 1L= 1.44 M glycine

x


3-42

Summary of mass-mole-number-volume

relationships in solution.

Figure 3.12MASS (g)

of compoundin solution

AMOUNT (mol)of compound

in solution


of compoundin solution

VOLUME (L)of solution

M (g/mol)

M (g/mol)Avogadro’s number(molecules/mol)


3-43

Sample Problem 3.14 Calculating Mass of Solute in a Given Volume of Solution

PROBLEM: A “buffered” solution maintains acidity as a reaction occurs. In living cells phosphate ions play a key buffering role, so biochemistry often study reactions in such solutions. How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate?

volume of soln

moles of solute

grams of solute

multiply by M

multiply by M

PLAN:

SOLUTION:

Molarity is the number of moles of solute per liter of solution. Knowing the molarity and volume leaves us to find the # moles and then the # of grams of solute. The formula for the solute is Na2HPO4.

1.75 L x0.460 moles

1 L

0.805 mol Na2HPO4 x 141.96 g Na2HPO4

mol Na2HPO4

= 0.805 mol Na2HPO4

= 114 g Na2HPO4


3-44

Laboratory preparation of molar solutions.Figure 3.12

A •Weigh the solid needed.•Transfer the solid to a volumetric flask that contains about half the final volume of solvent.

B Dissolve the solid thoroughly by swirling.

C Add solvent until the solution reaches its final volume.


3-45

Converting a concentrated solution to a dilute solution.Figure 3.13


3-46

Sample Problem 3.14Preparing a Dilute Solution from a Concentrated Solution

PROBLEM: “Isotonic saline” is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80 L of isotomic saline from a 6.0 M stock solution?

PLAN: It is important to realize the number of moles of solute does not change during the dilution but the volume does. The new volume will be the sum of the two volumes, that is, the total final volume.

volume of dilute soln

moles of NaCl in dilute soln = mol NaCl in concentrated soln

L of concentrated soln

multiply by M of dilute solution

divide by M of concentrated soln

MdilxVdil = #mol solute = MconcxVconc

SOLUTION:

0.80 L soln x

= 0.020 L soln

0.15 mol NaCl

L soln

0.12 mol NaCl x L solnconc

6 mol

= 0.12 mol NaCl


3-47

Sample Problem 3.15 Calculating Amounts of Reactants and Products for a Reaction in Solution

PROBLEM: Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide?

PLAN: Write a balanced equation for the reaction; find the grams of Mg(OH)2; determine the mol ratio of reactants and products;

use mols to convert to molarity.

mass Mg(OH)2

divide by M

mol Mg(OH)2

mol ratio

mol HCl

L HCl

divide by M


3-48

Sample Problem 3.15 Calculating Amounts of Reactants and Products for a Reaction in Solution

SOLUTION:

continued

Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)

0.10g Mg(OH)2 xmol Mg(OH)2

58.33g Mg(OH)2

= 1.7x10-3 mol Mg(OH)2

1.7x10-3 mol Mg(OH)2 x2 mol HCl

1 mol Mg(OH)2

= 3.4x10-3 mol HCl

3.4x10-3 mol HCl x 1L

0.10mol HCl= 3.4x10-2 L HCl


3-49

Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution

PROBLEM: Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Because of their toxicity, however, soluble mercury compounds, such mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide. How many grams of mercury(II) sulfide form?

PLAN: As usual, write a balanced chemical reaction. Since this is a problem concerning a limiting reactant, we proceed as we would for a limiting reactant problem. Find the amount of product which would be made from each reactant. Then choose the reactant that gives the lesser amount of product.


3-50

SOLUTION:

Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution

continued

L of Na2S

mol Na2S

mol HgS

multiply by M

mol ratio

L of Hg(NO3)2

mol Hg(NO3)2

mol HgS

multiply by M

mol ratio

Hg(NO3)2(aq) + Na2S(aq) HgS(s) + 2NaNO3(aq)

0.050 L Hg(NO3)2

x 0.010 mol/L

1 mol HgS

1 mol Hg(NO3)2

0.020 L Hg(NO3)2

x 0.10 mol/L

1 mol HgS

1 mol Na2S

= 5.0x10-4 mol HgS = 2.0x10-3 mol HgS

Hg(NO3)2 is the limiting reagent.

5.0x10-4 mol HgS x232.7g HgS

1 mol HgS= 0.12 g HgS

xx


3-51

Figure 3.15 An overview of the key mass-mole-number stoichiometry relationships.

Documents

Stoichiometry of Formulas and Equations ปริมาณสัมพันธ์ของสูตรและสมการ