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The Mole and Stoichiometry
H Chemistry IUnit 8
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Objectives #1-4 The Mole and its Use in Calculations
I. Fundamentals1 mole of an element or compound
contains 6.02 X 1023 particles and has a mass = to its molar mass
“The Triad of the Mole”1 mole = 6.02 X 1023 particles =
molar mass*henceforth the number 6.02 X 1023 will
be known as the Avogadro’s number
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*examples of mole quantities:1 mole of iron = 55.8 grams1 mole of copper =63.5 grams1 mole of water = 18.0 grams
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II. Introduction to Mole Problems(Follow the procedures outlined in Unit
1 for dimensional analysis problems)
1. Calculate the number of atoms in .500 moles of iron.
*Determine known:.500 atoms Fe
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*Determine unknown:atoms Fe*Use Triad of the Mole to determine
conversion factor:1 mole = 6.02 X 1023 atoms*Use conversion factor solve problem:.500 moles Fe X 6.02 X 1023 atoms/1
mole
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Check answer for units, sig. figs,, and reasonableness
3.01 X 1023 atoms Fe2. Calculate the number of atoms
in .450 moles of zinc..450 moles Zn X 6.02 X 1023 atoms
Zn /1 mole = 2.71 X 1023 atoms Zn
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3. Calculate the number of moles in 2.09 X 1023 atoms of sulfur.
2.09 X 1025 atoms S X 1mole S / 6.02 X 1023 atoms S = 34.7 moles S4. Calculate the number of moles in
3.06 X 1022 atoms of chlorine.3.06 X 1022 atoms Cl X 1 mole Cl / 6.02 X 1023 atoms Cl = .0508 moles Cl
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5. Calculate the number of atoms in 35.7 g of silicon.
35.7 g Si X 6.02 X 1023 atoms Si / 28.1 g
= 7.65 X 1023 atoms Si
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II. Molar Mass1. Calculate the molar mass of H3PO4:
3 H at 1.00 g = 3.00 g1 P at 31.0 g = 31.0 g4 O at 16.0 g = 64.0 gTotal = 98.0 g
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2. Calculate the molar mass of Al(OH)3:
1 Al at 27.0 g = 27.0 g3 O at 16.0 g = 48.0 g3 H at 1.0 g = 3.0 gTotal = 78.0 g
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3. Calculate the molar mass of BaCl2.2H2O:
1 Ba at 137.3 g = 137.3 g2 Cl at 35.5 g = 71.0 g2 H2O at 18.0 g = 36.0 g
Total = 244.2 g
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III.Mole-Mass Problems1. How many grams are in 7.20 moles
of dinitrogen trioxide?*determine known:7.20 moles N2O3
*determine unknown:grams N2O3
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*determine molar mass of compound if grams are involved:
76.0 g*use “Triad of the Mole” to determine
conversion factor:1 mole = 76.0 g*use conversion factor to solve problem:7.20 moles N2O3 X 76.0 g N2O3 / 1 mole =
547.2 g
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*check answer for units, sig. figs., and reasonableness:
547 g N2O3
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2. What is the mass in grams of 4.52 g of barium chloride?
4.52 g BaCl2 X 208.2 g / 1 mole = 941 g
3. Calculate the number of ammonium ions in 3.50 grams of ammonium phosphate.
3.50 g (NH4)3PO4 X
1 mole (NH4)3PO4 / 149. 0 g X
3 NH4+1 / 1 mole X
6.02 X 1023 ions / 1 mole NH4+1 =
4.24 X 1022 NH4+1 ions
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4. Calculate the mass of carbon in 7.88 X 1026 molecules of C8H18.
7.88 X 1026 molecules C8H18 X
1 mole C8H18 / 6.02 X 1023 molecules C8H18
X 8 moles C / 1 mole C8H18 X
12.0 g C / 1 mole C = 1.26 X 105 g C
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5. Calculate the number of molecules present in 2.50 moles of water.
2.50 moles H2O X 6.02 X 1023 molecules /
1 mole = 1.51 X 1024 molecules H2O
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6. Calculate the mass in grams of 4.50 X 1025 molecules of C6H10.
4.50 X 1025 molecules C6H10 X
82.0 g / 6.02 X 1023 molecules =6130 g C6H10
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7. Calculate the mass of 1 molecule of propane.
1 molecule C3H8 X
44.0 g / 6.02 X 1023 molecules = 7.31 X 10-23 g C3H8
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Objective #5 Characteristics of Solutions
Characteristic
Components solute and solvent
Particle Size .01 – 1 nm (atoms, ions, and molecules)
Tyndall Effect Result negative
Ability to be separated by filtration
no
Homogeneous or heterogeneous
homogeneous
Examples in different phases solid – brassliquid – saltwatergas - air
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Objective #5 Characteristics of Solutions
II. The Solution Process*dissociation and hydration of solutes:solute is split apart by solvent (dissociation)solute particles are surrounded by solvent
particles (hydration or solvation)*the rate of solution formation can be
increased by:stirring, raising temperature, and powdering
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Objective #5 Characteristics of Solutions
*behavior of ionic, polar, and nonpolar solutes in water:water is polar and will dissolve many ionic and polar solutes
NaCl → Na+1 + Cl-1
HCl → H+1 + Cl-1 nonelectrolytes vs. electrolytes
*”like dissolves like”: materials that have similar bonds will dissolve each other
NaCl (ionic) H2O (polar)
HCl (polar) H2O (polar)
I2 (nonpolar) CCl4 (nonpolar)
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Objective #5 Characteristics of Solutions
*miscible vs. immiscible liquids:miscible liquids dissolve in each other and form one
phase
immiscible liquids don’t dissolve in each other and form two phases
*unsaturated vs. saturated vs. supersaturated solutions:
Unsaturated (less solute dissolved than possible)Saturated (limit of solute dissolving)Supersaturated (beyond limit of solute dissolving)
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Objective #5 Characteristics of Solutions
*effect of pressure on solubility:gases will be more soluble in a liquid if the
atmospheric pressure is increased*effect of temperature on solubility:for solids in liquids – increasing temp. usually
increases solubilityfor gases in liquids – increasing temp.
decreases solubility (interpreting solubility graphs)
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Interpreting Solubility Graphs
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Objective #6 Molarity
*formula: Molarity = moles of solute / liter of
solution*examples:1. Calculate the molarity of a solution
prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.
11.5 g NaOH X 1 mole / 40.0 g = .288 molesM = .288 moles / 1.50 L = .192 M
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2. A chemist requires 1.00 L of .200 M potassium dichromate solution. How many grams of solid potassium dichromate will be required?
Moles = M X L = (.200 M) (1.00 L) = .200 moles K2Cr2O7
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.200 moles K2Cr2O7 X 294.2 g / 1 mole =
59 g K2Cr2O7
3. What is the molarity of each ion in the following solutions (assuming all are strong electrolytes)
.15 M calcium chloride:CaCl21 Ca to 2 Cl.15 M Ca+2
2(.15 M) = .30 M Cl-1
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.22 M calcium perchlorateCa(ClO4)2
1 Ca to 2 ClO4
.22 M Ca+2
2 (.22 M) = .44 M ClO4-1
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4. What is the molarity of an HCl solution made by diluting 3.50 L of a .200 M solution to a volume of 5.00 L?
moles before dilution = moles after dilution
recall moles = MVM1V1 = M2V2
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M1V1 = M2V2
#1 = concentrated#2 = dilutedM2 = M1V1 / V2
= (3.50 L) (.200 M) / 5.00 L = .140 M
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5. An experiment calls for 2.00 L of a .400 M HCl, which must be prepared from 2.00 M HCl. What volume in milliliters of the concentrated acid is needed to be diluted to form the .400 M solution?
M1V1 = M2V2
V1 = M2V2 / M1
= (.400 M) (2.00 L) / (2.00 M) = .400 L = 400. mlHow would one prepare solution?
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Objective #7 Percentage Composition
I. Formula% element in compound = mass of element in sample of
compound /mass of sample of compound
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*examples:1. Calculate the percentage composition
phosphate (K3PO4)
*determine molar mass of compound:212.4 g2. Determine total mass of each component:3 K at 39.1 g each = 117.3 g1 P at 31.0 g each = 31.0 g4 O at 16.0 g each = 64.0 g
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3. Divide total mass of each component by
molar mass of compound:% K = 117.3 g / 212.4 g = .55% P = 31.0 g / 212.4 g = .15% O = 64.0 g / 212.4 g = .30
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4. Multiply each result by 100 % and round appropriately.
.55 X 100 = 55 % K
.15 X 100 = 15 % P
.30 X 100 = 30 % O
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2. Calculate the percentage composition of sodium carbonate (Na2CO3)
molar mass = 106.0 g% Na = (46.0 g / 106.0 g) X 100 =
43%% C = (12.0 g / 106.0 g) X 100 = 11%% O = (48.0 g / 106.0 g ) X 100 = 45%
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Objective #8 Determining Empirical Formulas
*empirical vs. molecular formulas:H2O2 --› HO
H2O --› H2O
*examples:1. Determine the empirical formula for
a compound that when analyzed contained 70.9% potassium and 29.1% sulfur by mass.
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*If given percentages, assume 100 gram sample and convert percents to grams:
70.9% K --› 70.9 g K29.1% S --› 29.1 g S
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*convert masses to moles:70.9 g K X 1 mole K / 39.1 g = 1.81
moles K29.1 g S X 1 mole S / 32.1 g =.907
moles S*divide all mole values by the smallest
mole value in the set:.907 / .907 = 1 K1.81 / .907 =1.995 S
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*round resulting values to the nearest whole number or multiply by factor:
1 K2 S*use final values as subscripts and write
formula with the elements in order of increasing electronegativity; if compound is organic, list carbon first, then hydrogen, and the remainder by electronegativity:
K2S
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2. A compound of iron and oxygen when analyzed showed 70.0% iron and 30.0% oxygen by mass. Determine the empirical formula.
70. 0 g Fe X 1 mole/55.8 g = 1.25 moles Fe30.0 g O X 1 mole/16.0 g = 1.88 moles O1.25 / 1.25 = 11.88 / 1.25 = 1.51 Fe X 2 = 2 Fe 1.5 O X 2 = 3 OFe2O3
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3. Determine the empirical formula for a compound that contains 10.88 g of calcium and 19.08 g of chlorine.
10.88 g Ca X 1 mole / 40.1 g = .271 mole
19.08 g Cl X 1 mole / 35.5 g = .537 mole
.271 / .271 = 1 Ca
.537 / .271 = 2 ClCaCl2
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Objective #9 Determining Molecular Formulas
*examples:1. Analysis of a compound showed it to
consist of 80.0% carbon and 20.0% hydrogen by mass. The gram molecular mass is 30.0 g. Determine the molecular formula.
*determine empirical formula if needed:
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80.0 g C X 1 mole / 12.0 g = 6.67 moles C20.0 g H X 1 mole / 1.0 g = 20.0 moles H6.67 / 6.67 = 1 C20.0 / 6.67 = 3 HCH3
*determine empirical formula mass (efm):15 g
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*if efm matches gram molecular mass (gmm), then empirical formula is the same as molecular
15 g ≠ 30 g*if masses don’t match, divide gmm by efm
to determine factor:30 g / 15 g = 2*multiply subscripts of empirical formula by
factor to obtain molecular formula:C2H6
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2. If the empirical formula for a compound is C2HCl, determine the molecular formula if the gram molecular mass is 181.5 g.
emp. mass 60.5 g181.5 g / 60.5 g = 2C4H2Cl2
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3. A compound contains 58.5% carbon, 9.8% hydrogen, 31.4% oxygen and the gram molecular mass is 102 g. Determine the molecular formula.
58.5 g C X 1 mole / 12.0 g = 4.88 mole C9.8 g H X 1 mole / 1.0 g = 9.8 mole H31.4 g O X 1 mole / 16.0 g = 1.96 mole O4.88 / 1.96 = 2.5 C9.8 / 1.96 = 5 H1.96 / 1.96 = 1 O
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2.5 C X 2 = 5 C5 H X 2 = 10 H1 O X 2 = 2 Oemp. formula = C5H10O2
emp. mass = 102 gmolecular formula = C5H10O2
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Objectives #10-11 Introduction to Stochiometry
*Stoichiometry is a method of calculating amounts in a chemical reaction
I. Interpreting Chemical Equations Quantitatively
*example: 4Al + 3O2 --› 2Al2O3
*the following information can be determined from this reaction:
1. Number of particles2. Number of moles3. Mass
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1. Particles:4Al + 3O2 --› 2Al2O3
atom molecule formula unit4 atoms Al3 molecules O2
2 formula units Al2O3
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2. Moles4Al + 3O2 --› 2Al2O3
4 moles Al3 moles O2
2 moles Al2O3
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3. Mass:4Al + 3O2 --› 2Al2O3
108 g Al96 g O2
204 g Al2O3
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4Al + 3O2 --› 2Al2O3
*playing with the mole ratios:4 moles Al --› ? Al2O3
4 moles Al --› 2 moles Al2O3
8 moles Al --› ? Al2O3
8 moles Al --› 1 mole Al2O3
2 moles Al --› ? Al2O3
2 moles Al --› 1 mole Al2O3
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4Al + 3O2 --› 2Al2O3
.200945 moles Al --› ?
.200945 moles Al X 2 moles Al2O3 / 4 mole Al =
.100473 mole Al
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Objective #12 Stoichiometric Calculations
*examples:1. If 20.0 g of magnesium react with
excess hydrochloric acid, how many grams of magnesium chloride are produced?
*write balanced chemical equation if not provided:
Mg + 2HCl --› MgCl2 + H2
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*determine known:20.0 g*determine unknown:g MgCl2*convert given to moles:20.0 g Mg X 1 mole Mg / 24.3 g *select and utilize appropriate mole-mole to
convert to moles of unknown:X 1 mole MgCl2 / 1 mole Mg
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*if unknown is to be measured in moles skip to step 7. If unknown is to be measured in grams, multiply by unknown’s molar mass. If unknown is to be measured in particles, multiply by Avogadro’s number:
X 95.3 g MgCl2 / 1 mole MgCl2
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*perform math:78.4 g MgCl2*check answer for units, sig figs, and
reasonableness
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2. How many moles of water can be formed from 7.5 moles of ethyne gas (C2H2) reacting with excess oxygen gas?
2C2H2 + 5O2 --› 4CO2 + 2H2O
7.5 moles C2H2 X
2 moles H2O / 2 moles C2H2 =
7.5 moles H2O
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3. If 50.00 g of Rb reacts with excess S8 how many formula units of Rb2S can be formed?
16 Rb + S8 --› 8Rb2S
50.00 g Rb X 1 mole Rb / 85.5 g Rb X8 mole Rb2S / 16 mole Rb X
6.02 X 1023 f. units Rb2S / 1 mole Rb2S
= 1.760 X 1023 f. units Rb2S
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Objective #13 Limiting Reagents
*General Terms:excess reagent reactant that is leftoverlimiting reagent reactants that is used up*Examples:1. How many grams of carbon dioxide can be
obtained by the action of 50.0 g of sulfuric acid on 100.0 g of calcium carbonate?
(follow general steps for stoichiometry, just do it twice)
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H2SO4 + CaCO3 --› CaSO4 + CO2 +H2O
50.0 g H2SO4 X 1 mole H2SO4 / 98.1 g X
1 mole CO2 / 1 mole H2SO4 X
44.0 g CO2 / 1 mole CO2 = 22.4 g CO2
------------------------------------------100.0 g CaCO3 X 1 mole CaCO3 / 100.1 g X
1 mole CO2 / 1 mole CaCO3 X
44.0 g CO2 / 1 mole CO2 = 43.96 g CO2
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(examine two answers produced; the smaller of the two is the right answer of product produced)
*the reactant quantity that provides the smaller correct answer is called the limiting reagent
*the reactant quantity that provides the larger incorrect answer is called the excess reagent
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2. How many grams of hydrogen gas can be produced by the reaction of 100. g of phosphoric acid on 25.0 g of aluminum?
2H3PO4 + 2Al --› 2AlPO4 + 3H2
100. g H3PO4 X 1 mole H3PO4 / 98.0 g X
3 mole H2 / 2 mole H3PO4 X
2.0 g H2 / 1 mole H2 = 3.06 g H2
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25.0 g Al X 1 mole Al / 27.0 g Al X3 mole H2 / 1 mole H2 X 2.0 g H2 / 1
mole=2.78 g H2
2.78 g H2 (smaller value)
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Determine how much of the excess reagent is left over.
*determine limiting reagent:Al*use stochiometry to convert from mass of
limiting reagent to find mass of excess reagent actually used:
25.0 g Al X 1 mole Al / 27.0 g Al X2 mole H3PO4 / 2 mole Al X
98.0 g H3PO4 / 1mole = 90.7 g H3PO4 used
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*Subtract amount of excess reagent used from original amount of excess reactant given in problem to determine mass of excess reactant left over:
100. g – 90.7 g = 9 g left over
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3. How many grams of sodium chloride can be produced from then reaction of 15.0 g of chlorine gas and 15.0 g of sodium bromide in the above reaction? Determine how much of the excess reagent is left over?
Cl2 + 2NaBr --› Br2 + 2NaCl
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15.0 g Cl2 X 1 mole Cl2 / 71.0 g Cl2 X
2 mole NaCl / 1 mole Cl2 X
58.5 g NaCl / 1 mole NaCl = 24.7 g NaCl
--------------------------------------------15.0 g NaBr X 1 mole NaBr / 102.9 g
NaBrX 2 mole NaCl / 2 mole NaBr X58.5 g NaCl / 1 mole NaCl = 8.53 g
NaCl
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15.0 g NaBr X 1 mole NaBr / 102.9 g NaBr
X 1 mole Cl2 / 2 mole NaBr X
71.0 g Cl2 / 1 mole Cl2 = 5.17 g Cl2 (used)
--------------------------------------------15.0 g – 5.17 g = 9.8 g Cl2 leftover
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Objective #15 Percentage Yield
*General Formula:Percent Yield = (actual yield / theoretical yield) X 100%*examples:1. When 9.00 g of aluminum react with an
excess of phosphoric acid, 30.0 g of aluminum phosphate are produced. What is the percentage yield of this reaction?
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*write balanced chemical equation:2Al + 2 H3PO4 --› 3H2 + 2AlPO4
*determine actual yield of a product in reaction given in the problem:
30.0 g AlPO4
*use given reactant quantity and general stochiometry procedures to determine mass of appropriate product:
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9.00 g Al X 1 mole Al / 27.0 g Al X2 mole AlPO4 / 2 mole Al X
122.0 g / 1 mole = 40.7 g AlPO4
*divide actual yield from problem by the theoretical yield calculated to find percentage yield:
%Y = (30.0 g / 40.7 g) X 100 = 74%
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2. Calculate the percentage yield if 6.00 g of aluminum are produced from the decomposition of 25.0 g of aluminum oxide.
2Al2O3 --› 4Al + 3O2
25.0 g Al2O3 X 1 mole Al2O3 / 102.0 g X
4 mole Al / 2 mole Al2O3 X
27.0 g Al / 1 mole Al = 13.2 g Al%Y= (6.00 g / 13.2 g) X 100% = 45%