Stoichiometry of Formulas and Equations ปริมาณสัมพันธ์ของสูตรและสมการ Chapter 3 Stoichiometry of Formulas and Equations ปริมาณสัมพันธ์ของสูตรและสมการ

Mole - Mass Relationships in Chemical Systems 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 3.5 Fundamentals of Solution Stoichiometry

The Mole 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)

Counting objects of fixed relative mass. Figure 3.1 Counting objects of fixed relative mass. 12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 1023 atoms Fe 32.07g S = 6.022 x 1023 atoms S

One mole of common substances. Figure 3.2 Oxygen 32.00 g One mole of common substances. Water 18.02 g CaCO3 100.09 g Copper 63.55 g

Table 3.1 Summary of Mass Terminology Definition Unit Isotopic mass Mass of an isotope of an element amu amu Atomic mass Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance (also called atomic weight) Molecular (or formula) mass (also called molecular weight) Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) amu Molar mass (M) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units) g/mol (also called gram-molecular weight)

Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Table 3.2 Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Atoms/mole of compound 6(6.022 x 1023) atoms 12(6.022 x 1023) atoms 6(6.022 x 1023) atoms Mass/molecule of compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu Mass/mole of compound 72.06 g 12.10 g 96.00 g

Interconverting Moles, Mass, and Number of Chemical Entities Mass (g) = no. of moles x no. of grams 1 mol g No. of moles = mass (g) x no. of grams 1 mol M No. of entities = no. of moles x 6.022x1023 entities 1 mol No. of moles = no. of entities x 6.022x1023 entities 1 mol

Summary of the mass-mole-number relationships for elements. Figure 3.3 MASS(g) of element M (g/mol) AMOUNT(mol) of element Avogadro’s number (atoms/mol) ATOMS of element

Sample Problem 3.1 = 3.69 g Ag = 1.04x1024 atoms Fe Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element Sample Problem 3.1 PROBLEM: (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? PLAN: (a) To convert mol of Ag to g we have to use the #g Ag/mol Ag, the molar mass M. amount(mol) of Ag mass(g) of Ag multiply by M of Ag (107.9g/mol) mol Ag 107.9 g Ag = 3.69 g Ag SOLUTION: 0.0342 mol Ag x PLAN: (b) To convert g of Fe to atoms we first have to find the #mols of Fe and then convert mols to atoms. mass(g) of Fe amount(mol) of Fe atoms of Fe divide by M of Fe (55.85g/mol) 55.85g Fe mol Fe SOLUTION: 95.8g Fe x = 1.72 mol Fe 6.022x1023atoms Fe mol Fe multiply by 6.022x1023 atoms/mol 1.72mol Fe x = 1.04x1024 atoms Fe

Summary of the mass-mole-number relationships for compounds. Figure 3.3 MASS(g) of compound M (g/mol) chemical formula AMOUNT(mol) of elements in compound AMOUNT(mol) of compound Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound

number of (NH4)2CO3 formula units amount(mol) of (NH4)2CO3 divide by M Calculating the Moles and Number of Formula Units in a Given Mass of a Compound Sample Problem 3.2 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? PROBLEM: PLAN: 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. mass(g) of (NH4)2CO3 number of (NH4)2CO3 formula units amount(mol) of (NH4)2CO3 divide by M multiply by 6.022x1023 formula units/mol SOLUTION: 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 mol (NH4)2CO3 96.09 g (NH4)2CO3 6.022x1023 formula units (NH4)2CO3 mol (NH4)2CO3 x 41.6 g (NH4)2CO3 x = 2.61x1023 formula units (NH4)2CO3

Mass percent from the chemical formula Mass % of element X = atoms of X in formula x atomic mass of X (amu) x 100 molecular (or formula) mass of compound (amu) Mass % of element X = moles of X in formula x molar mass of X (amu) molecular (or formula) mass of compound (amu) x 100

(a) What is the mass percent of each element in glucose? Calculating the Mass Percents and Masses of Elements in a Sample of Compound Sample Problem 3.3 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? amount(mol) of element X in 1mol compound PLAN: We have to find the total mass of glucose and the masses of the constituent elements in order to relate them. multiply by M (g/mol) of X mass(g) of X in 1mol of compound SOLUTION: divide by mass(g) of 1mol of compound Per mole glucose there are 6 moles of C 12 moles H 6 moles O (a) mass fraction of X multiply by 100 mass % X in compound

Sample Problem 3.3 Calculating the Mass Percents and Masses of Elements in a Sample of Compound continued 1.008 g H mol H 12.01 g C mol C 12 mol H x = 12.096 g H 6 mol C x = 72.06 g C 16.00 g O mol O M = 180.16 g/mol 6 mol O x = 96.00 g O (b) 72.06 g C 180.16 g glucose mass percent of C = = 0.3999 x 100 = 39.99 mass %C 12.096 g H 180.16 g glucose mass percent of H = = 0.06714 x 100 = 6.714 mass %H 96.00 g O 180.16 g glucose mass percent of O = = 0.5329 x 100 = 53.29 mass %O

Empirical and Molecular Formulas Empirical 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. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the empirical formula.

NaClO4 is sodium perchlorate. Determining the Empirical Formula from Masses of Elements Sample Problem 3.4 PROBLEM: 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? PLAN: 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). mol Na 22.99 g Na SOLUTION: 2.82 g Na x = 0.123 mol Na mass(g) of each element mol Cl 35.45 g Cl divide by M(g/mol) 4.35 g Cl x = 0.123 mol Cl amount(mol) of each element mol O 16.00 g O use # of moles as subscripts 7.83 g O x = 0.489 mol O preliminary formula Na1 Cl1 O3.98 Na1 Cl1 O3.98 NaClO4 change to integer subscripts NaClO4 is sodium perchlorate. empirical formula

Determining a Molecular Formula from Elemental Analysis and Molar Mass Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: 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. PLAN: assume 100g lactic acid and find the mass of each element divide each mass by mol mass(M) amount(mol) of each element molecular formula use # mols as subscripts preliminary formula divide mol mass by mass of empirical formula to get a multiplier convert to integer subscripts empirical formula

C3H6O3 is the molecular formula Determining a Molecular Formula from Elemental Analysis and Molar Mass Sample Problem 3.5 continued SOLUTION: Assuming there are 100 g of lactic acid, the constituents are 40.0 g C x mol C 12.01g C 6.71 g H x mol H 1.008 g H 53.3 g O x mol O 16.00 g O = 3.33 mol C = 6.66 mol H = 3.33 mol O C3.33 H6.66 O3.33 CH2O empirical formula 3.33 3.33 3.33 mass of CH2O molar mass of lactate 90.08 g 30.03 g 3 C3H6O3 is the molecular formula

CnHm + (n+ ) O2 = n CO(g) + H2O(g) m 2 Figure 3.4 Combustion train for the determination of the chemical composition of organic compounds. CnHm + (n+ ) O2 = n CO(g) + H2O(g) m 2

Determining a Molecular Formula from Combustion Analysis Sample Problem 3.6 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? difference (after-before) = mass of oxidized element PLAN: find the mass of each element in its combustion product preliminary formula empirical formula molecular formula find the mols

Molecular formular = C6H8O6 Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis continued SOLUTION: CO2 85.35 g - 83.85 g = 1.50 g H2O 37.96 g - 37.55 g = 0.41 g 12.01 g CO2 44.01 g CO2 There are 12.01 g C per mol CO2 1.50 g CO 2 x = 0.409 g C 2.016 g H2O 18.02 g H2O There are 2.016 g H per mol H2O 0.41 g H2O x = 0.046 g H 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 176.12 g/mol 88.06 g C1H1.3O1 C3H4O3 = 2.000 Molecular formular = C6H8O6

Table 3.3 Some Compounds with Empirical Formula CH2O (Composition by Mass: 40.0% C, 6.71% H, 53.3% O) Molecular Formula Whole-Number Multiple M (g/mol) Name Use or Function formaldehyde CH2O 1 30.03 disinfectant; biological preservative acetic acid C2H4O2 2 60.05 acetate polymers; vinegar (5%soln) lactic acid C3H6O3 3 90.09 sour milk; forms in exercising muscle erythrose C4H8O4 4 120.10 part of sugar metabolism ribose C5H10O5 5 150.13 component of nucleic acids and B2 glucose C6H12O6 6 180.16 major energy source of the cell CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6

Table 3.4 Two Pairs of Constitutional Isomers C4H10 C2H6O Property Butane 2-Methylpropane Ethanol Dimethyl Ether M(g/mol) 58.12 58.12 46.07 46.07 Boiling Point -0.50C -11.060C 78.50C -250C Density at 200C 0.579 g/mL (gas) 0.549 g/mL (gas) 0.789 g/mL (liquid) 0.00195 g/mL (gas) Structural formulas Space-filling models

The formation of HF gas on the macroscopic and molecular levels. Figure 3.6

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

Balancing Chemical Equations translate the statement balance the atoms adjust the coefficients check the atom balance specify states of matter

Balancing Chemical Equations Sample Problem 3.7 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. PLAN: SOLUTION: translate the statement C8H18 + O2 CO2 + H2O balance the atoms C8H18 + O2 CO2 + H2O 25/2 8 9 adjust the coefficients 2C8H18 + 25O2 16CO2 + 18H2O check the atom balance 2C8H18 + 25O2 16CO2 + 18H2O specify states of matter 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g)

Summary of the mass-mole-number relationships in a chemical reaction. Figure 3.8 MASS(g) of compound A MASS(g) of compound B M (g/mol) of compound A M (g/mol) of compound B AMOUNT(mol) of compound A molar ratio from balanced equation AMOUNT(mol) of compound B Avogadro’s number (molecules/mol) Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound A MOLECULES (or formula units) of compound B

write and balance equation 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 mols Cu2O find g SO2 find mols O2 find kg O2

Sample Problem 3.8 Calculating Amounts of Reactants and Products continued SOLUTION: 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? 3mol O2 2mol Cu2S 10.0mol Cu2S x (a) = 15.0mol O2 (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? 10.0mol Cu2S x 2mol SO2 2mol Cu2S 64.07g SO2 mol SO2 (b) x = 641g SO2 (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? 2.86kg Cu2O x 103g Cu2O kg Cu2O mol Cu2O 143.10g Cu2O (c) x = 20.0mol Cu2O = 0.959kg O2 kg O2 103g O2 20.0mol Cu2O x 3mol O2 2mol Cu2O 32.00g O2 mol O2 x x

Writing an Overall Equation for a Reaction Sequence 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: SOLUTION: write balanced equations for each step 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) 2Cu2O(s) + 2C(s) 4Cu(s) + 2CO(g) cancel reactants and products common to both sides of the equations sum the equations 2Cu2S(s)+3O2(g)+2C(s) 4Cu(s)+2SO2(g)+2CO(g)

An ice cream sundae analogy for limiting reactions. Figure 3.10 An ice cream sundae analogy for limiting reactions.

Table 3.5 Information Contained in a Balanced Equation Viewed in Terms of Reactants C3H8(g) + 5O2(g) Products 3CO2(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

Using Molecular Depictions to Solve a Limiting-Reactant Problem Sample Problem 3.10 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. (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? PLAN: 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.

Sample Problem 3.10 continued 3.00 mol F2 0.750 mol Cl2 139 g ClF3 Using Molecular Depictions to Solve a Limiting-Reactant Problem continued SOLUTION: Cl2(g) + 3F2(g) 2ClF3(g) (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). F2 Cl2 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. Since we find that the ratio is 4/1, that means F2 is in excess and Cl2 is the limiting reactant. 3.00 mol F2 0.750 mol Cl2 = 4 1 2 mol ClF3 1 mol Cl 92.5 g ClF3 1 mol ClF3 0.750 mol Cl2 x x = 139 g ClF3

Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant Sample Problem 3.11 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. mass of N2H4 mass of N2O4 limiting mol N2 divide by M multiply by M mol of N2H4 mol of N2O4 g N2 molar ratio mol of N2 mol of N2

Sample Problem 3.11 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant continued SOLUTION: 2 N2H4(l) + N2O4(l) N2(g) + H2O(l) 3 4 mol N2H4 32.05g N2H4 1.00x102g N2H4 x N2H4 is the limiting reactant because it produces less product, N2, than does N2O4. = 3.12 mol N2H4 3 mol N2 2mol N2H4 3.12mol N2H4 x = 4.68mol N2 mol N2 28.02g N2 4.68mol N2 x = 131g N2 mol N2O4 92.02g N2O4 2.00x102g N2O4 x = 2.17mol N2O4 3 mol N2 mol N2O4 2.17mol N2O4 x = 6.51mol N2

A + B (reactants) C (main product) D (side products) Figure 3.11 The effect of side reactions on yield. A + B (reactants) C (main product) D (side products)

Sample Problem 3.12 40.10 g SiC mol SiC 103g 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: SOLUTION: write balanced equation SiO2(s) + 3C(s) SiC(s) + 2CO(g) 103 g SiO2 kg SiO2 mol SiO2 60.09 g SiO2 find mol reactant & product 100.0 kg SiO2 x x = 1664 mol SiO2 mol SiO2 = mol SiC = 1664 find g product predicted 40.10 g SiC mol SiC kg 103g 1664 mol SiC x x = 66.73 kg actual yield/theoretical yield x 100 51.4 kg 66.73 kg percent yield x 100 =77.0%

Calculating the Molarity of a Solution 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? PLAN: Molarity is the number of moles of solute per liter of solution. mol of glycine SOLUTION: divide by volume 0.715 mol glycine 495 mL soln 1000mL 1 L x concentration(mol/mL) glycine 103mL = 1L = 1.44 M glycine molarity(mol/L) glycine

mass-mole-number-volume relationships in solution. Figure 3.12 MASS (g) of compound in solution Summary of mass-mole-number-volume relationships in solution. M (g/mol) AMOUNT (mol) of compound in solution Avogadro’s number (molecules/mol) M (g/mol) MOLECULES (or formula units) of compound in solution VOLUME (L) of solution

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? PLAN: 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. volume of soln SOLUTION: multiply by M 0.460 moles 1 L moles of solute 1.75 L x = 0.805 mol Na2HPO4 multiply by M 141.96 g Na2HPO4 mol Na2HPO4 0.805 mol Na2HPO4 x grams of solute = 114 g Na2HPO4

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. C Add solvent until the solution reaches its final volume. B Dissolve the solid thoroughly by swirling.

Converting a concentrated solution to a dilute solution. Figure 3.13 Converting a concentrated solution to a dilute solution.

MdilxVdil = #mol solute = MconcxVconc Preparing a Dilute Solution from a Concentrated Solution Sample Problem 3.14 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. MdilxVdil = #mol solute = MconcxVconc volume of dilute soln multiply by M of dilute solution SOLUTION: moles of NaCl in dilute soln = mol NaCl in concentrated soln 0.15 mol NaCl L soln 0.80 L soln x = 0.12 mol NaCl divide by M of concentrated soln L solnconc 6 mol 0.12 mol NaCl x = 0.020 L soln L of concentrated soln

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 HCl L HCl divide by M mol Mg(OH)2 mol ratio

Sample Problem 3.15 Calculating Amounts of Reactants and Products for a Reaction in Solution continued SOLUTION: Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) mol Mg(OH)2 58.33g Mg(OH)2 0.10g Mg(OH)2 x = 1.7x10-3 mol Mg(OH)2 2 mol HCl 1 mol Mg(OH)2 = 3.4x10-3 mol HCl 1.7x10-3 mol Mg(OH)2 x 1L 0.10mol HCl 3.4x10-3 mol HCl x = 3.4x10-2 L HCl

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.

Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution continued 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) SOLUTION: 0.050 L Hg(NO3)2 0.020 L Hg(NO3)2 L of Na2S mol Na2S mol HgS multiply by M mol ratio x 0.010 mol/L x 0.10 mol/L 1 mol HgS 1 mol Hg(NO3)2 1 mol HgS 1 mol Na2S x x = 5.0x10-4 mol HgS = 2.0x10-3 mol HgS Hg(NO3)2 is the limiting reagent. 232.7g HgS 1 mol HgS 5.0x10-4 mol HgS x = 0.12 g HgS

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