141
Summer Chemistry 2008
Silberberg Chapter 3
Dr. Ben Tovrog
3-1
Silberberg Chapter 3 Problem Assignment
These problems are for your practice. They will not be collected or graded.
Easy and Medium: 3.1, 3.2, 3.6, 3.7, 3.8, 3.10, 3.12, 3.14, 3.18, 3.19, 3.20, 3.25, 3.27, 3.33, 3.37, 3.39, 3.41, 3.44, 3.46, 3.51, 3.53, 3.55, 3.61, 3.63, 3.65, 3.67, 3.69, 3.71, 3.73, 3.75, 3.77, 3.79, 3.83, 3.85, 3.92, 3.94, 3.96, 3.98, 3.102, 3.105, 3.116, 3.118, , 3.120, 3.124, 3.128 Hard: 3.137, 3.139, 3.142, 3.151, 3.155
3-2
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
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
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
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Figure 3.2
Oxygen 32.00 g
One mole of common substances.
CaCO3 100.09 g
Water 18.02 g
Copper 63.55 g
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Table 3.1
Term Isotopic mass
Summary of Mass Terminology
Definition Mass of an isotope of an element Unit amu
Atomic mass
(also called atomic weight) Molecular (or formula) mass (also called molecular weight)
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)
amu
amu
Molar mass (M) Mass of 1 mole of chemical entities (also called (atoms, ions, molecules, formula units) gram-molecular weight)
g/mol
3-7
Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol)
Table 3.2
Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Carbon (C) Hydrogen (H)
Oxygen (O)
6 atoms 6 moles of atoms 6(6.022 x 1023) atoms
12 atoms 12 moles of atoms 12(6.022 x 1023) atoms
6 atoms 6 moles of atoms 6(6.022 x 1023) atoms
Mass/molecule of compound
Mass/mole of compound
6(12.01 amu) =72.06 amu
72.06 g
12(1.008 amu) =12.10 amu
12.10 g
6(16.00 amu) =96.00 amu
96.00 g
3-8
Interconverting Moles, Mass, and Number of Chemical Entities no. of grams Mass (g) = no. of moles x
g
1 mol
1 mol
No. of moles = mass (g) x
M
no. of grams 6.022x1023 entities 1 mol 1 mol
No. of moles = no. of entities x
No. of entities = no. of moles x
6.022x1023 entities
3-9
Moles The Central Concept
Volume of Gas (L)
22.4 liters @STP
X 6.02 x 10
23
X 22.4 liters @STP
X molar mass
Number of molecules
Mole
23
6.02 x 10
molar mass X Volume (L)
Mass (g)
Volume (L)
Molarity (mole/Liter)
Shown for molecules. For atoms substitute atomic mass for molar mass
3-10
Mole Problems
In one dozen elephants, how many trunks are there? 12 How many legs? 48 How many methane molecules are there in one mole of methane? 6.022 E 23
How many carbon atoms are there in one mole of methane CH4 ? _6.022 E 23 C atoms_
How many hydrogen atoms are there in one mole of methane CH4? (6.022 E 23)x4 = 2.41 E 24 How many carbons atoms are there in 0.23 moles of methane? (6.022 E 23) x 0.23 = 1.38 E 23 How many H atoms are there in 0.23 moles of methane? (2.41 E 24) x .23
How many total atoms (C and H) are there in 0.23 moles of methane? 6.9 E 25
What is the mass of one mole of C atoms (e.g. diamonds, or graphite)? 12.01 g (recall that 1 amu is equalivent to 1.6606 x 10 -24g). What is the mass of one mole of CH4 molecules? 12.01 + 4X(1.008) = 16.04 What is the mass (in grams) of a single CH4 molecule? 16.04 / 6.022 E 23 = 2.66 E -23 g
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Sample Problem 3.1 PROBLEM:
Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element
(a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe?
a) 0.0342 mol X 107.9 g/mol = 3.69 g Ag b) 95.8g Fe = 1.72mol Fe x 6.02E 23 = 1.04E24 atoms Fe 55.85 g/mol
3-12
Sample Problem 3.2 PROBLEM:
Calculating the Moles and Number of Formula Units in a Given Mass of a Compound
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?
3-13
Mole Within the Mole
Ethanol has a molecular formula of CH3CH2OH. What is the mass of 1.00 mole of ethanol?
What is the mass of 0.5623 moles of ethanol? How many moles of ethanol are present in a 100.0g sample of ethanol? How many moles of each element (C, H, O) are present in a 100.0 g samples of ethanol?
How many grams of each element (C, H, O) are present in a 100.0 g sample of ethanol?
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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) x 100
molecular (or formula) mass of compound (amu)
3-15
Sample Problem 3.3
Calculating the Mass Percents and Masses of Elements in a Sample of 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? PLAN: We have to find the total mass of glucose and the masses of the constituent elements in order to relate them. amount(mol) of element X in 1mol compound multiply by M(g/mol) of X
SOLUTION: (a)
Per mole glucose there are 6 moles of C 12 moles H 6 moles O
mass(g) of X in 1mol of compound divide by mass(g) of 1mol of compound mass fraction of X
multiply by 100
mass % X in compound
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Sample Problem 3.3
continued 12.01 g C mol C 6 mol O x 16.00 g O
Calculating the Mass Percents and Masses of Elements in a Sample of Compound
6 mol C x
= 72.06 g C
12 mol H x
1.008 g H mol H
= 12.096 g H
= 96.00 g O
M = 180.16 g/mol
mol 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
mass percent of H = 180.16 g glucose 96.00 g O = 0.06714 x 100 = 6.714 mass %H
mass percent of O =
180.16 g glucose
= 0.5329 x 100 = 53.29 mass %O
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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.
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Sample Problem 3.4
Determining the Empirical Formula from Masses of Elements
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?
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). SOLUTION: 2.82 g Na mol Na = 0.123 mol Na 22.99 g Na mass(g) of each element mol Cl divide by M(g/mol) 4.35 g Cl = 0.123 mol Cl 35.45 g Cl amount(mol) of each element mol O 7.83 g O = 0.489 mol O use # of moles as subscripts 16.00 g O preliminary formula Na1 Cl1 O3.98 NaClO4 change to integer subscripts empirical formula NaClO4 is sodium perchlorate. PLAN:
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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.
Assume: 100g Lactic Acid .... So: 40g Carbon / 12.01 = 3.33 mol 6.71g Hydrogen / 1.008 = 6.66 mol 2 53.3g Oxygen / 16.00 = 3.33 mol 1 ratio 1 Empirical = CH2O
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Sample Problem 3.5
Determining a Molecular Formula from Elemental Analysis and Molar Mass
3-21
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
Name
M (g/mol) 30.03 60.05
Use or Function
disinfectant; biological preservative
acetate polymers; vinegar(5% soln) sour milk; forms in exercising muscle
formaldehyde
acetic acid
CH2O
C2H4O2
1 2 3 4 5
lactic acid
erythrose ribose
C3H6O3
C4H8O4
90.09
120.10 150.13 180.16
part of sugar metabolism
component of nucleic acids and B2
C5H10O5
C6H12O6
glucose
6
major energy source of the cell
CH2O
C2H4O2
C3H6O3
C4H8O4
C5H10O5
C6H12O6
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translate the statement
balance the atoms
adjust the coefficients
check the atom balance
specify states of matter
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Table 3.5 Information Contained in a Balanced Equation Viewed in Terms of
molecules
Reactants C3H8(g) + 5O2(g)
1 molecule C3H8 + 5 molecules O2
Products 3CO2(g) + 4H2O(g)
3 molecules CO2 + 4 molecules H2O
amount (mol) mass (amu) mass (g) total mass (g)
1 mol C3H8 + 5 mol O2
44.09 amu C3H8 + 160.00 amu O2 44.09 g C3H8 + 160.00 g O2 204.09 g
3 mol CO2 + 4 mol H2O
132.03 amu CO2 + 72.06 amu H2O 132.03 g CO2 + 72.06 g H2O
204.09 g
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Sample Problem 3.7 PROBLEM:
Balancing Chemical Equations
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.
3-25
Balancing Equations
Cr (s) + S8 (s) Cr2S3(s)
2.
NaHCO3 (s)
Na2CO3 (s)
+
CO2 (g)
+
H2O (g)
Fe2S3 (s)
+
HCl (g)
FeCl3 (s) +
H2S (g)
4.
CS2 (l)
(s)
+
NH3 (g)
H2S (g)
+
NH4SCN
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Figure 3.6
The formation of HF gas on the macroscopic and molecular levels.
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Sample Problem 3.8
PROBLEM:
Calculating Amounts of Reactants and Products
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?
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Sample Problem 3.8
continued SOLUTION:
Calculating Amounts of Reactants and Products
2Cu2S(s) + 3O2(g)
2Cu2O(s) + 2SO2(g)
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Calculating Amounts of Reactants and Products
The thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rockets. The reaction is:
Fe2O3(s)
+
2 Al(s)
2 Fe (l)
+
Al2O3 (s)
What masses of iron (III) oxide and aluminium must be used to produce 15.0 g of iron? What mass of aluminium oxide would be produced?
Fe2O3(s)
______ g
+
2 Al(s)
______ g
2 Fe (l)
15 g
+
Al2O3 (s)
_______ g
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Figure 3.4
Combustion train for the determination of the chemical composition of organic compounds.
CnHm + (n+
m m ) O2 = n CO2(g) + H2O(g) 2 2
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Sample Problem 3.6
NEED TO
PROBLEM:
Determining a Molecular Formula from Combustion Analysis STUDY THIS!!!!
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 mass of CO2 absorber before combustion mass of H2O absorber after combustion mass of H2O absorber before combustion =85.35g =83.85g =37.96g =37.55g
What is the molecular formula of vitamin C?
PLAN: difference (after-before) = mass of oxidized element
find the mass of each element in its combustion product
preliminary empirical formula formula molecular formula
find the mols
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Sample Problem 3.6
continued SOLUTION:
Determining a Molecular Formula from Combustion Analysis
NEED TO STUDY THIS!!!!
H2O 37.96 g-37.55 g = 0.41 g 12.01 g CO2
44.01 g CO2
CO2
85.35 g-83.85 g = 1.50 g
There are 12.01 g C per mol CO2
1.50 g CO2
= 0.409 g C
There are 2.016 g H per mol H2O. 0.41 g H O 2.016 g H2O 2 18.02 g H2O O must be the difference: 0.409 g C 12.01 g C C1H1.3O1 C3H4O3 = 0.0341 mol C 1.000 g - (0.409 + 0.046) = 0.545 0.046 g H 1.008 g H 176.12 g/mol 88.06 g = 0.0456 mol H
= 0.046 g H
0.545 g O 16.00 g O
= 0.0341 mol O
= 2.000
C6H8O6
3-33
Cheese sandwich analogy for limiting reactions.
In a reaction of two defined amounts of reactants one reactant will be the "limiting reagent" and the other the "reagent in excess." In order to find the limiting reagent, determine the reactant, which if totally consumed generates the least amount of product. This reactant is the limiting reagent.
2 Bread slices
+ 1 cheese slice
1 sandwich.
A. If you were given 8 slices of bread, and 4 slices of cheese, how many sandwiches could you make? What (if any) "reactants" would be left over?
B. If you were given 8 slices of bread and 2 slices of cheese, how many sandwiches could you make? What (if any reactants) would be left over?
C. If you were given 6 slices of bread and 4 slices of cheese, how many sandwiches could you make? What (if any reactants) would be left over?
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Sample Problem 3.10 Using Molecular Depictions to Solve a LimitingReactant 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. The stoichiometry of the reaction is 3 moles flourine reacts with one mole chlorine.
(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.
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Sample Problem 3.10 Using Molecular Depictions to Solve a LimitingReactant Problem
SOLUTION: Cl2(g) + 3F2(g) 2ClF3(g)
F2
Cl2
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Sample Problem 3.11 PROBLEM:
Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant
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 divide by M mass of N2O4 limiting mol N2 multiply by M
mol of N2H4
molar ratio mol of N2
mol of N2O4
g N2
mol of N2
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Sample Problem 3.11
continued SOLUTION:
Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant
2 N2H4(l) + N2O4(l) mol N2H4 32.05g N2H4
3 N2(g) + 4 H2O(l)
1.00x102g N2H4
= 3.12mol N2H4
3.12mol N2H4
3 mol N2
N2H4 is the limiting reactant because it produces less product, N2, than does N2O4.
28.02g N2 mol N2
= 4.68mol N2
2mol N2H4 2.00x102g N2O4
mol N2O4 92.02g N2O4 2.17mol N2O4
4.68mol N2 = 2.17mol N2O4
= 131g N2
3 mol N2
mol N2O4
= 6.51mol N2
3-38
Limiting Reagents
Freon-12 (CCl2F2) is synthesized by the reaction between carbon tetrachloride and antimony trifluoride at a temperature slightly above room temperature. Balance the following chemical equation (Hint: Balance F first.): CCl4(l) + ___ SbF3(s) __ CCl2F2(g) + ___ SbCl3(s)
Suppose 5.0 mol of antimony trifluoride is added to 10.0 mol of carbon tetrachloride. Which is the limiting reagent?
How many moles of each species (CCl4, SbF3, CCl2F2, and SbCl3) remain after the reaction is completed (assume 100% completion).
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Figure 3.11
The effect of side reactions on yield.
A +B
(reactants)
C
(main product)
D
(side products)
3-40
Yield Definitions
Theoretical Yield The amount of product predicted based on the stoichiometry of the reaction. It assumes 100 % perfect reaction conversion and product purity.
Actual Yield The amount of product actually obtained.
Actual Yield
Percent Yield =
Theoretical Yield
X 100%
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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? SOLUTION:
PLAN:
write balanced equation
find mol reactant & product
SiO2(s) + 3C(s) 100.0 kg SiO2
103 g SiO2 kg SiO2
SiC(s) + 2CO(g)
mol SiO2
60.09 g SiO2
= 1664 mol SiO2
mol SiO2 = mol SiC = 1664
find g product predicted
1664 mol SiC
actual yield/theoretical yield x 100
40.10 g SiC kg mol SiC 103g
= 66.73 kg
percent yield
51.4 kg 66.73 kg
x100 =77.0%
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Percent Yields
Magnesium burns rapidly in air to form magnesium oxide. This reaction gives off an enormous amount of energy in the form of light and is used in both flares and fireworks. If 13.5 g of MgO are formed when 10.0 grams of magnesium are burned in air, what is the percent yield of the reaction? Balance the following reaction: Mg (s) + O2 (g) MgO(s)
What is the theoretical yield of MgO(s) ? What is the percent yield of MgO(s)?
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Molarity
When two substances are mixed together and the resulting mixture is homogeneous, we call the mixture a solution. The component of a solution in the greater amount is the solvent, and the component in the lesser amount is the solute
moles of A Molarity of A =
Units mol/Liter
Volume (Liters)
moles of A = (Molarity of A) * (Volume)
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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
divide by volume concentration(mol/mL) glycine 103mL = 1L molarity(mol/L) glycine
SOLUTION:
0.715 mol glycine 495 mL soln 1000mL
= 1.44 M glycine
1L
3-45
Molarity
1. A 20-oz (591 mL) bottle of soft drink has 66.5 g of sugar (C12H22O11, molar mass = 342.3 g/mol). What is the molarity of sugar in the solution?
(66.5g) / (342.3 g/mol) = 0.194 mol
M = (0.19 mol) / (0.591 L) = 0.329 M
2. You are given a 250 mL solution of sodium sulfate Na2 SO4 which is 0.150 M solution. a. How many moles of sodium sulfate are contained in the solution?
a) mole = M x V = (0.15 mol/L) (0.250 L) = 0.0375 mole
b. How many grams of sodium sulfate are contained in the solution?
0.0375 mole (142.05g/mol) = 5.33 g
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Moles The Central Concept
Volume of Gas (L)
+ 22.4 liters @STP
X 6.02 x 10
23
X 22.4 liters @STP
X molar mass
Number of molecules
Mole
23
6.02 x 10
molar mass X Volume (L)
Mass (g)
Volume (L)
Molarity (mole/Liter)
Shown for molecules. For atoms substitute atomic mass for molar mass
3-47
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?
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. SOLUTION:
PLAN:
volume of soln
multiply by M moles of solute multiply by M grams of solute
1.75 L 0.460 moles
1L
= 0.805 mol Na2HPO4
0.805 mol Na2HPO4 141.96 g Na2HPO4
mol Na2HPO4
3-48
= 114 g Na2HPO4
Figure 3.12
Laboratory preparation of molar solutions.
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.
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Figure 3.13
Converting a concentrated solution to a dilute solution.
3-50
Sample Problem 3.14 Preparing 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?
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 SOLUTION: multiply by M of dilute solution moles of NaCl in dilute soln = mol NaCl in concentrated soln divide by M of concentrated soln L of concentrated soln 0.80 L soln 0.15 mol NaCl = 0.12 mol NaCl L soln 0.12 mol NaCl L solnconc = 0.020 L soln 6 mol
PLAN:
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Molarity, Dilutions
1. Concentrated hydrochloric is 12.1 M. If you are instructed to prepare 350.0 mL of a 0.975 M HCl solution, how many milliliters of the concentrated HCl solution should you dispense in preparing the dilute solution?
2. 100.0 mL of a 0.050 M solution of NaCl in water is diluted to 250.0 mL. What is the final concentration of the solution?
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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
L HCl
divide by M mol HCl
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Sample Problem 3.15 Calculating Amounts of Reactants and Products for a Reaction in Solution
SOLUTION:
Mg(OH)2(s) + 2HCl(aq)
MgCl2(aq) + 2H2O(l)
3-54
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.
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Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution
SOLUTION: Hg(NO3)2(aq) + Na2S(aq) HgS(s) + 2NaNO3(aq)
L of Hg(NO3)2 multiply by M
mol Hg(NO3)2 mol ratio
mol HgS
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