unit.03.stoichiometry1.formulas&equations.lecture
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8/12/2019 Unit.03.Stoichiometry1.Formulas&Equations.lecture
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Stoichiometry
Unit # 3Stoichiometry:
Calculations with ChemicalFormulas and Equations
CHM 1045: General Chemistry and
Qualitative Analysis
Dr. Jorge L. Alonso
Miami-Dade College – Kendall Campus
Miami, FL
Textbook Reference:
•Module #3 & 4
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Stoichiometry
CO2 (g)H2O(g)
O2 (g)
CH4 (g)
Chemical
Reaction
The actual
phenomenon that
occurs when chemicalinteract with each other.
Methane gas is mixed with air and thenit is light-up by a spark.
flame
What is happening here?
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Stoichiometry
(1) Decomposition:
(2) Combination (Synthesis):
(3) Double Displacement (Replacement) or Metathesis,
Exchange
(4) Single Displacement (Replacement)
(5) Combustion
AB + CD AD + CB where A & C are Metals, B & D Nonmetals
MN + M
MN + N
M
or
N
MN +
AB A + B
A + B AB
: reactions of oxygen with an organic compounds
(hydrocarbons, alcohols) that produce CO2 + H2O and a flame.
C3H8 (g ) + 5 O2 (g ) 3 CO2 (g ) + 4 H2O (g )
Predicting Products: Types of ReactionsWhat happens when substances react?
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Stoichiometry
• Sodium + Chlorine
• Dihydrogen Monoxide
• Magnesium + Hydrochloric Acid
• Hydrochloric Acid + Calcium Hydroxide
• Combustion (burning with oxygen) of:
Sucrose (C12H22O11)
Octane
Chemical Equations
What happens when you mix (cause a reaction of) the following?
Write balanced chemical equations for each.
Sodium Chloride
Hydrogen + oxygen
Magnesium Chloride + Hydrogen
Hydrogen hydroxide + Calcium chloride
+ Oxygen Carbon dioxide + water
+ Oxygen Carbon dioxide + water
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Stoichiometry
Predicting Products, writing Formulas
and Balancing Equations
Na + Cl2
H2O Mg + HCl
HCl + Ca(OH)2
C12H22O11 + O2
C8H18 + O2
NaCl
H2 + O2 MgCl2 + H2
HOH + CaCl2
CO2 + H2O
CO2 + H2O
2
2 22
2 2
12 11
8 912.52 25 16 18
2
12
*
• Sodium + Chlorine
• Dihydrogen Monoxide
• Magnesium + Hydrochloric Acid
• Hydrochloric Acid + Calcium Hydroxide
Combustion (burning with oxygen) of:Sucrose (C12H22O11)
Octane
Sodium Chloride
Hydrogen + oxygen
Magnesium Chloride + Hydrogen
Hydrogen hydroxide + Calcium chloride
Carbon dioxide + water
Carbon dioxide + water
Alonso’s Rules for BE: (1) Easy element 1st hard
elements last.
(2) One element at a time.
(3) Use fractions when
necessary.
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Stoichiometry
Decomposition Reactions
{AirBags Movie*}
sodium azide (N31-) ∆
Simple: Binary compounds break down into their
constituent elements
2H2O 2H2 + O2
2NaCl(l) 2Na (l) + Cl2(g)
2NaN3(s) 2Na(s) + 3N2(g)
electrolysis
electrolysis
heat
2H2O2 2H2O + O2
Catalyst
{Peroxide Movie}
Important Exception:
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Stoichiometry
Decomposition Reactions
CaCO3 (s) CaO (s) + CO2 (g )
∆ 2 KClO3 (s) 2 KCl (s) + 3O2 (g )
Chlorates break down to metal chlorides and oxygen
Complex Compounds decompose into simpler compounds
2 NaOH (aq)
Na2
O (s)
+ H2
O (l )
2 H3PO4 (aq) P2O5(g ) + 3H2O (l )
Acids break down to nonmetal oxides and water
Bases break down to metal oxides and water
2HNO3 (aq) N2O5(g ) + H2O (l )
All carbonates break down to metal oxides and carbon dioxide
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Stoichiometry
Ammonium carbonate powder is heated strongly
Na2O + CO2 + H2O
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Stoichiometry
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Stoichiometry
Combination (Synthesis) Reactions
2 Mg (s) + O2 (g ) 2 MgO (s)
Zn (s)
+ S (s)
ZnS (s)
2 H2 (g ) + O2 (g ) 2 H2O (l )
2 Al (s) + 3 Br 2 (l ) 2 AlBr 3 (s)
Simple:
• Two or more elements react to form one compound
A + B AB
{Mg Movie}
{H2O Movie*}
{AlBr 3 Movie*}
{ZnS Movie*}
Now let’s balance equations
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Stoichiometry
Bromine liquid is poured over aluminum metal
Hydrogen chloride and ammonia gas are mixed together.
Sulfur dioxide gas is bubbled into water.
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Stoichiometry
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Stoichiometry
Metathesis (Double Displacement)
Example: what quantity of Baking Soda will react with 100mL of vinegar?
NaHCO3 (s) + HC2H3O2 (l ) NaC2H3O2 (aq) + HHCO3 (aq)
H2CO3 (aq) H2O (l ) + CO2 (g ) (2nd Rx decomposition)
• Involve two Compounds
• Elements (or polyatiomic
groups) in the twocompounds exchange
partners
{Movie: Bicarb + Vineg with Stoichio&LimitReag *}
AB + CD AD + CB where A & C are Metals, B & D Nonmetals
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Stoichiometry
Metathesis (Double Displacement):
Acid-Base Neutralization Reaction
Examples:
HCl (aq ) + NaOH (aq ) NaCl (aq ) + HOH (l )
2 HCl (aq ) + Ca(OH)2(aq ) CaCl2 (aq ) + 2 HOH (l )
HCl (aq ) + NH4OH (aq ) NH4Cl (aq ) + 2 HOH (l )
Acid: compound containing hydrogen and a non metal (HN)
Bases: a metal hydroxide (MOH)
HN + MOH MN + HOH Acid + Base Salt + Water
{Movie: A-B RxNo Ind} {Movie: A-B Rx Ind+pH meter*}
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Stoichiometry
Single Displacement Reactions
A more active element displacing a less active elementsfrom a compound.
MN + M
MN + N
Activity (Electromotive) Series:
Metals: Li > K > Ba > Sr > Ca > Na > Mg > Al > Zn > Cr > Fe >
Cd > Co > Ni > Sn > Pb > (H) > Sb > Bi > Cu > Hg > Ag > Pd >
Pt > Au
Halogens: F >Cl > Br > I
M
or
N
(Single Replacement Rx.)
MN +
compound
element
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Stoichiometry
Single Displacement Reactions
• Examples:
Cu (s) + 2 AgNO3 (aq) Cu (s) + Zn(NO3)2 (aq)
Cl2 (g) + 2 NaBr (aq)
Activity (Electromotive) Series:
Metals: Li > K > Ba > Sr > Ca > Na > Mg > Al > Zn > Cr > Fe >
Cd > Co > Ni > Sn > Pb > (H) > Sb > Bi > Cu > Hg > Ag > Pd >
Pt > Au
Halogens: F >Cl > Br > I
{Movie:Cu+AgNO3}
A more active element displacing a less active
elements from a compound.
2 Ag + Cu(NO3)2 (aq)
2 NaCl (aq) + Br 2 (aq)
No Reaction
Activity series can also be found in
form of Reduction Potential table.
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Stoichiometry
Most
Active
Nonmetal
Most
Active
Metal
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Stoichiometry
H+OH-
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Stoichiometry
Reactions with Oxygen
(2) Combustion Reaction: Rapid reactions of
oxygen with an organic compounds
(hydrocarbons, alcohols) that produce CO2 + H2O
and a flame.
Examples:
CH4 (g ) + 2 O2 (g ) CO2 (g ) + 2 H2O (g )
C3H8 (g ) + 5 O2 (g ) 3 CO2 (g ) + 4 H2O (g )
Oxidation Rx.
Combustion Rx
(1) Oxidation Reactions: are combination
reactions involving oxygen.
{Movie: Mg, Fe, P, S + conc. O2
{Metal Oxides*}
What is the difference?
{Movie: CH3OH + O2*}
Wh i
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Stoichiometry
*When oxygen is scarce….
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Stoichiometry
gram-Molar Mass (g-MM)= Atomic Weigh, Formula
Weigh or Molecular Weight
Mass : Weight (in grams)
Mole = 6.022 x 1023 particles
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Stoichiometry
gram-Molar Mass (g-MM): AW, FW, MW
the mass in grams of 1 mole ofa substance (units= g/mol)
For an element we find it on the
periodic table.
For compounds the same as the formula & molecular weight
(but in g/mol)
Example: the g-MM of Al2(SO4)3, would be
2 Al: 2x(26.98 amu) = 53.96
+ 3 S: 3x(32.06 amu) = 96.18
+3x4 O: 12x(16.00 amu) =192.00
342.14 amu (g/mol)
*
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Stoichiometry
Formula Weight (FW)
• Sum of the atomic weights for the atoms in a chemical formula unit
(ionic compound)• So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1x(40.1 amu)
2 x Cl: 2x(35.5 amu)
111.1 amu
Molecular Weight (MW)
• Sum of the atomic weights of the atoms in a molecule (covalent
compound)
• For the molecule ethane, C2H6, the molecular weight would be
2 x C: 2x(12.0 amu)
6 x H: 6x(1.0 amu)
30.0 amu
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Stoichiometry
(Mass) Percent Composition
Percentage mass of a element (Na) in compound (NaCl):
100 xwhole
part %
Problem: (1) calculate mass % Au in Nagyagite.
(2) If you buy 1 kg of the ore, how much gold
does it have?
% element =
(# atoms of element) (atomic weight of Au)
(MW of Pb5 Au(TeSb)4S5) x 100
Nagyagite Gold Ore:
% Au =(1) ( 197)
( 2,301)
x 100 = 8.56 %
Oreg1000Auof g?
Ore g
Au g
100
56.8 Au g 5.68
100x NaClmass
Namass Na%
Pb5 Au(TeSb)4S5
g-MM = 2,301g/
40%100x58g
23g
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Stoichiometry
The Mole ConceptDermatological BiologicalChemical
Avogadro's Number:6.022,141,410,704,090,840,990,72 x 1023
602,214,141,070,409,084,099,072 .
602 sextillion
thousandmillionbilliontrillionquadrillionpentillionsextillion
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Stoichiometry
Using Equivalences as Mole Ratios:
MM-g
mole1
particles10x6.022
mole123
mole1
MM-g
mole1
particles10x6.02223
or
g-MM = 1 Mole ( ) = 6.022 x 10 23 particles
NaCl = 58g/η (Atoms or molecules)
From Equivalences we obtain useful Ratios or Conversion factors:
or
particles10 x6.02223
g-MM
g-MM
particles10 x6.02223
or
*
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Stoichiometry
Mole Calculations:
g-MM Moles # of Particles
mol0.0548
g187
NaClmol1 g.44358
g-MM Moles:
Moles # of Particles:
? g = 3.20 mol of NaCl
? mol = 3.20 g of NaCl
? f.u. = 3.2 mol NaCl
? mol = 3.2 x 10 52 f.u. NaCl
g.44358
NaClmol1
or
f.u.10x023.6
NaClmol1
23
NaClmol1
f.u.10x023.6 23
or
g.44358
NaClmol1
NaClmol1 g.44358
f.u.10x023.6
NaClmol1
23
NaClmol1
f.u.10x023.6 23
*
Which ratios will you need?
= 5.3 x 10 28 mol
= 1.9 x 10 24 f.u.
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Stoichiometry
Mole Calculations:
g-MM Moles # of Particles
f.u.10x6.022
g58.44323
g-MM # of Particles:
? g = 4.2 x 10 34 f. u. of NaCl
? f. u. of NaCl = 3.2 g of NaCl
f.u.10x6.022mol1
23
4.1 x 1012 g
g58.443
mole13.3 x 1022 f.u.
4.1 x 1012 g
g58.443
f.u10x6.02223
3.3 x 1022 f.u.
mol1g58.443
mole1
f.u.10x6.02223
*M l C l l i
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Stoichiometry
233 )(PHFe(CO)of f.u.1
atoms15
? atoms = 0.50 mole Fe(CO)3(PH3)2
mole1
f.u.10x6.02223
= 4.5 x 1024 atoms
*
How many molecules of H2O in 29g of water? How many atoms?
O gH molecules 229?
mole
molecules
1
10022.623
O gH
mole
218
1molecules
23102.2
O H of moleculesatoms 223 102.2? atoms10x6.6
23
O H moleculeatoms
21 3
Mole Calculations:
g-MM Moles # of Particles(atoms or molecules)
1+ 3+ 3+ 2+ 6 =
= 3.0 X 1023 f.u.
15
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Stoichiometry
The Determination of
Empirical Formulas ofCompounds by
Elemental Analysis
CxHy
Combustion
furnace
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Stoichiometry
Types of Formulas
• Empirical formulas give the lowestwhole-number ratio of atoms of eachelement in a compound.
• Molecular formulas give the exactnumber of atoms of each element in acompound.
Why are empirical formulas needed?
• Structural formulas (skeletal or space-
filling) show the order in which atoms arebonded and their three-dimensional shape.
Benzene, C6H6
HO CH H2O
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Stoichiometry
Elemental Analyses
Compounds are broken downand the masses of their constituent
elements are measured. From
these masses the empirical
formulas can be determined.
Expt. Data: (68g) 4g H
64g O
EmpF
HO
How do we determine the formula of a compound?
H4H1
1
g
mole
O4O16
1
g
mole
H14
O14
Mole Ratio
HxOy
moles
*
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Stoichiometry
Calculating Empirical Formulas
Calculate the empirical formula (mole ratio) from
the percent composition (% mass).
Problem:The compound para-aminobenzoic acid (you may have seen it
listed as PABA on your bottle of sunscreen) is composed of
carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and
oxygen (23.33%). Find the empirical formula of PABA.
*
carbon (61.31%),hydrogen (5.14%),
nitrogen (10.21%),
oxygen (23.33%).
carbon (61.31g),hydrogen (5.14g),
nitrogen (10.21g),
oxygen (23.33g)
Percent means out of 100, so assume a
100g sample of the compound, then….
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Stoichiometry
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
? mol C = 61.31 g x = 5.105 mol C
? mol H = 5.14 g x = 5.09 mol H
? mol N = 10.21 g x = 0.7288 mol N
? mol O = 23.33 g x = 1.456 mol O
1 mol
12.01 g
1 mol
14.01 g
1 mol
1.01 g
1 mol
16.00 g
What is the smallest mole ratio of the elements in this compound?
= 7.005 7
= 6.984 7
= 1.000
= 2.001 2
5.105 mol
0.7288 mol
5.09 mol
0.7288 mol0.7288 mol
0.7288 mol1.458 mol
0.7288 mol
Calculate the mole ratio by dividing by the smallest number of moles.
These are the subscripts for the empirical formula: C7H7NO2
61% C
5% H
10% N
23% O
Combustion Analysis: i th d f
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Stoichiometry
Combustion Analysis: is a method of
experimentally determining empirical formulas
• Compounds containing C, H and O are routinely analyzed through combustion in a
chamber like this
– C is determined from the mass of CO2 produced
– H is determined from the mass of H2O produced
– O is determined by difference after the C and H have been determined
{Movie}CxHy
Combustion
furnace
Magnesium
perchlorate
Sodium
hydroxide
Cx
Hy
+ O2
CO2
+ H2
O
mass
mass of C?
mass of H2O
2COg44.011
Cg12.011
OHg18.0
Hg1.0
2
mass of CO2
mass of H?
mass of H2Omass of CO2
How do you calculate the mass of C
in CO2 and that of H in H2O?
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Stoichiometry
CxHy (g ) + O2 (g ) CO2 (g ) + H2O (g )
Calculating Empirical Formulas
A 5.00 g sample of an unknown hydrocarbon was burned and produced
14.6 g of CO2. What is the empirical formula of the unknown
compound?
2COg44.011
Cg12.011? g C = 14.6 g CO2 = 3.98 g C
g H = 5.00 CxHy – 3.98 g C = 1.02 g H
? mol C = 3.98 g C
? mol H = 1.02 g H
g12.011
Cmole1
g1.011
Hmole1
= 0.332 mol C
= 1.01 mol H
/ 0.332 = 1.00
/ 0.332 = 3.04
EmpiricalFormula
CH3
5.00 g 14.6 g
(12g:32g=44g)
2006 A
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Stoichiometry
2006 A
?g C =
?g N =
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Stoichiometry
2003 B
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Stoichiometry
2003 B
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Stoichiometry
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Stoichiometric Calculations
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Stoichiometry
OHmol2
Omol1
2
2
Omol1
Hmol2
2
2
OHmol2
Hmol2
2
2
Mole Ratios from Balanced Equation:
The coefficients in the balanced equation can also be
interpreted as mole ratios of reactants and products
Stoichiometric Calculations2 2
Stoichiometric Calculations *
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Stoichiometry
Stoichiometric Calculations
mole ratio from balanced equation
MgOg2.35 O g? 2
MgOg2.35 O g? 2
MgOg40.304
MgOmol1
MgOmol2
Omol1 2
2
2
Omol1
Og32.000 2Og0.933
2 Mg (s ) + O2 (g ) 2 MgO (s )
2 Mg (s ) + O2 (g ) 2 MgO (s )
gramsNo direct
calculation
grams
Change:1. grams of MgO mol MgO
2. mol of MgO mol O2
3. mol of O2 grams of O2
How many grams of O2 are
required to form 2.35 g of MgO?
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St i hi t Li iti R t t
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Stoichiometry
In this example the sugar would be the limiting reactant,
because it will limit the amount of cookies you can make.
{ MovieLimitingReactants: Zn + 2 HCl ZnCl2 + H2 }
Stoichiometry: Limiting Reactants
Make cookies until you run out ofone of the ingredients
(or, too much of one reactant and not enough of the other)
8/12/2019 Unit.03.Stoichiometry1.Formulas&Equations.lecture
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Limiting & Excess Reactants
8/12/2019 Unit.03.Stoichiometry1.Formulas&Equations.lecture
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Stoichiometry
Limiting & Excess Reactants
Problem
If 5.0g of both Mg and O2 are used:(1) Which is the limiting and the excess
reactants?
(2) How much of the excess will be leftunreacted ?
(3) How much MgO will be produced ?
2 Mg (s ) + O2 (g ) 2 MgO (s )
Limiting & Excess Reactants
8/12/2019 Unit.03.Stoichiometry1.Formulas&Equations.lecture
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Stoichiometry
Limiting & Excess Reactants
Problem
? g O2 = 5.0 g Mg
? g Mg = 5.0 g O2
2
2
Omol1
g32.0
Mgmol2
Omol1
g24.3
Mgmol1
2 Mg (s ) + O2 (g ) 2 MgO (s )
?g MgO = 5.0 g Mg
Mgmol1
g24.3
Omol1
Mgmol2
g32.0
Omol1
2
2
MgOmol1
g40.3
Mgmol2
MgOmol2
g24.3
Mgmol1
MgOg8.29
= 3.29 g O2
= 7.59 g Mg
Mg is limiting reactant and O2 is excess reactant
Which one do I use to determine the MgO produced by rx?
5 g 5 g
8/12/2019 Unit.03.Stoichiometry1.Formulas&Equations.lecture
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Limiting Reactant Experiments
8/12/2019 Unit.03.Stoichiometry1.Formulas&Equations.lecture
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Stoichiometry
Experiments:
0.0025 g + 0.0050 g
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
Limiting Reactant Experiments
gZn gHCl 0025.0?
gHCl gZn 0050.0?
Zn35.453g
1 Znmole
Znmole
H mole
1
Cl2
Cl1
Cl40.66
H mole
H g
gHCl 0093.0
HClg66.40
Cl1 H mole
Cl2
n1
H mole
Z mole
n1
Zn453.35
Z mole
g
gZn0013.0
HCl is Limiting Reactant
Zn is Excess Reactant
How many grams of ZnCl2 are produced in this reaction?
gHCl gZnCl 0050.0? 2
HClg66.40
Cl1 H mole
Cl2
nCl1 2
H
Z
2
2
nCl1
ZnCl233.166
Z mole
g 20063.0 gZnCl
How many grams of ZnCl2 are produced in this reaction?
8/12/2019 Unit.03.Stoichiometry1.Formulas&Equations.lecture
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Stoichiometry
Theoretical Yield
the amount of product that can be made as
calculated by stoichiometry.
the amount reaction actually produces (less)
Actual Yield
Percent Yield
Actual Yield
Theoretical YieldPercent Yield = x 100
gHCl gZnCl 0050.0? 2
HClg66.40
Cl1 H mole
Cl2
nCl1 2
H
Z
2
2
nCl1
ZnCl233.166
Z mole
g 20063.0 gZnCl
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
20045.0BalanceUsing gZnCl
%4.711000063.
0045.
*
8/12/2019 Unit.03.Stoichiometry1.Formulas&Equations.lecture
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Stoichiometry
HGeFg160. 3
The following reaction has a 95% yield:
GeH4 + 3GeF4
4GeF3Hg-MM: 76.622 148.5756 130.58
Problem:
How many grams of the product are formed, when 23.4 g of GeH4 are
reacted with excess GeF4?
?g GeF3H = 23.4 g GeH4 H GeF
g 130.58
GeH 1
H GeF 4
g 76.622
GeH
34
34
1
1
HGeFof g152 0.95H)GeFg(160. 33
Since % yield is only 95%, then actual yield is:
What would the yield be if reaction Actually produced 130g only?
100160
130%
g
g Yield %81
100
120%75
whole
g
100
%75
120 g whole g 160