Download unit iv – stoichiometry 1

UNIT IV – STOICHIOMETRY
1
* Stoichiometry - the study of quantitative relationships in chemical formulas and equations
* Chemical Formulas - a series of symbols and numbers used to represent the composition of an element
or a compound
I. Formulas for Compounds
A. Ionic Compounds - represents ratio of cations to anions
* remember, ionic compounds don’t form molecules as we think of them
* we call “molecules” formula units when we deal with ionic compounds
- +
* one unit of this ratio (
) is referred to as a formula unit rather than a
molecule
+2
* EXAMPLE: CaCl2 means that for every one Ca ion in the compound,
-
there are two Cl ions
* Many ionic compounds have a fixed ratio of water
molecules in their crystal structure, these are called hydrates;
these water molecules need to be included in the formula
.
* EXAMPLE: BaSO4 8 H2O means that for each formula
unit of barium hydroxide, the crystal structure has 8
molecules of water imbedded inside
B. Molecular Compounds - represents actual number of atoms of that element per molecule of
that compound
* EXAMPLE: CO2 means that each CO2 molecule contains one C
atom and two O atoms
II. Formula Mass - a measure of the mass in one mole of molecules (or formula units) of that substance
- a.k.a.: molecular mass, or molar mass
16.0 amu
O
H
1.0 amu
1.0 amu
1.0 amu
+ 16.0 amu
18.0 amu
H
1.0 amu
* units for these calculations are atomic mass units (amu) = grams
per mole (g/mol)
* The atomic mass of carbon is ~12.0 g/mol. This means that one
23
mole (6.02 x 10 ) atoms of carbon will weigh about 12.0 g.
* The formula mass of carbon dioxide (CO2) is ~44.0 g/mol. This
23
means that one mole (6.02 x 10 ) of carbon dioxide molecules
will weigh about 44.0g.
UNIT IV – STOICHIOMETRY
Find the formula mass of the following:
S.
1) glucose: C6H12O6
2) Al2(SO4)3
3) CoCl2-6 H2O
4) cisplatin: Pt(NH3)2Cl2
6(12) + 12(1.0) + 6(16.0) = 18θ g/mol
(58.9) + 2(35.5) + 6(18.0) = 238 g/mol
2(27.0) + 3(32.1) + 12(16.0) = 342 g/mol
(195) + 2(17.0) + 2(35.5) = 30θ g/mol
III. Moles – 1 mole of anything is equivalent to 6.022045 x 1023 of that thing
* in chemistry the mole is a number used in conjunction with atoms and molecules
1 mole = 12.0 g 12C = 6.022045 x 1023 atoms/molecules/formula units
* We can “get to” moles through a variety of different methods
A. Mass  moles
* use the calculated formula mass for the element or compound
S.
1. How many moles of helium atoms are in 6.46g of He?
6.46 g He !
1 mol He
= 1.62 mol He
4.00 g He
2. How many grams are there in 0.356 moles of zinc chloride?
0.356 mol ZnCl 2 !
136 g ZnCl 2
= 48.5 g ZnCl 2
1 mol ZnCl 2
B. Molecules/Atoms/Formula units  moles
1. How many molecules of CH4 are there in 0.256 moles of the gas?
0.256 mol CH 4 !
6.02 x 10 23 CH 4
= 1.54 x 10 23 mcs CH 4
1 mol CH 4
2. If you have a sample containing 9.45 x 1023 formula units of NaCl, how many moles is that?
9.45 x 10 23 fu NaCl !
1 mol NaCl
= 1.57 mol NaCl
6.02 x 10 23 fu NaCl
* You can also go easily from mass to atoms/molecules/formula units by going “through” moles
1. How many atoms are there in 16.3g of sulfur (S)?
16.3 g S !
1 mol S 6.02 x 10 23 at S
!
= 3.06 x 10 23 at S
32.1 g S
1 mol S
2. How many grams of will 5.36 x 1022 formula units of NaF weigh?
5.36 x 10 22 fu NaF !
1 mol NaF
42.0 g NaF
!
= 3.74 g NaF
23
6.02 x 10 fu NaF 1 mol NaF
2
UNIT IV – STOICHIOMETRY
3
C. Moles and Gases
* represented by “n” in the Ideal Gas Law: PV = nRT  solve for n:
n=
PV
RT
R = 0.0821 L-atm / mol-K = 8.31 L-kPa / mol-K
T must be in Kelvin: K = oC + 273.15
* if conditions are Standard Temperature and Pressure (STP – 0oC & 1 atm pressure)
1 mole of gas occupies 22.4 liters of space
S. Assume conditions are at STP:
1) How many moles of He gas occupy 44.8 L of space?
44.8 L He !
1 mol He
= 2.00 mol He
22.4 L He
2) If you have 3.77 mole of H2 gas, how much space will it occupy?
3.77 mol H 2 !
22.4 L H 2
= 84.4 L H 2
1 mol H 2
S. Assume conditions are not standard:
1) How many moles of CH4 gas occupy 67.2 L of space at a pressure of 0.982 atm and a temperature
of 19.0oC?
n=
PV
=
RT
(0.982atm)(67.2 L)
= 2.75 mol CH 4
atm • L
(0.0821
)(292.15 K )
mol • K
2) How many molecules of CH4 gas are present in a 10.0 L container at a pressure of 1.05 atm and
a temperature of 25.0o C?
n=
PV
=
RT
6.02 x 10 23 mcs CH 4
(1.05atm)(10.0 L)
= 0.429 mol CH 4 !
= 2.58 x 10 23 mcs CH 4
atm • L
1 mol CH 4
(0.0821
)(298.15 K )
mol • K
3) How many grams of helium gas would you have in a 2.50 L balloon at 100.3 kPa of pressure and
a temperature of 21.0o C?
n=
PV
=
RT
(100.3kPa)(2.50 L)
4.00 g He
= 0.10258 mol He !
= 0.410 g He
kPa • L
1 mol He
(8.31
)(294.15 K )
mol • K
UNIT IV – STOICHIOMETRY
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IV. Equation Conversion Problems
N2(g) + 3 H2(g)  2 NH3(g)
1 mole molecules + 3 moles molecules 
CANNOT SAY:
1 g N2
+
3 g H2

2 moles molecules
2 g NH3
* As a result, when you want to get from grams of one quantity in the equation to grams of
another quantity, you need to go “through” moles
A. Mole-mole Problems (one-step)
S.
1) How many moles of NH3 are formed from 4.56 moles of N2?
4.56 mol N 2 !
2 mol NH 3
= 9.12 mol NH 3
1 mol N 2
2) If you begin with 4.50g of nitrogen, how many grams of hydrogen are needed to complete the reaction?
4.50 g N 2 !
1 mol N 2 3 H 2 2.02 g H 2
!
!
= 0.974 g H 2
28 g N 2 1 N 2 1 mol H 2
3) If you begin with 4.50g of nitrogen, how many grams of ammonia do you expect?
4.50 g N 2 !
1 mol N 2 2 NH 3 17 g NH 3
!
!
= 5.46 g NH 3
28 g N 2
1 N 2 1 mol NH 3
4) If you begin this reaction with 100.0 g of nitrogen gas at a pressure of 1.03 atm and a temperature
of 23.0oC, and all of the nitrogen gas is consumed in the reaction, what volume should the
ammonia gas produced by the reaction occupy?
1 mol N 2 2 NH 3
!
= 7.14 mol NH 3
28 g N 2
1 N2
nRT (7.14mol )(0.0821)(296.15 K )
V=
=
= 169 L
P
1.03atm
100.0 g N 2 !
V. Percent Yield - a means of measuring the “success” of a reaction
* - Using stoichiometry, we are predicting the amount of a product we EXPECT. In a real
laboratory situation, however, the amount of product you get often differs. The percent
yield gives a numerical value which represents the percent of the amount expected you
got in the experiment.
Percent Yield =
Actual Amount
x100%
Expected Amount
S. If the experiment in problem #3 above yielded 5.00g of ammonia, what is the percent yield?
5.00 g
x100% = 91.6%
5.46 g
UNIT IV – STOICHIOMETRY
5
VI. Limiting Reactant Problems
* all previous problems assumed we had as much reactant as we needed  NOT REAL
* example: burning wood:
wood + O2  CO2(g) + H2O(l) + other hydrocarbons + ash
- limiting reactant – the reactant consumed first by the reaction; as a result, it determines:
1) how much of the product will be formed
2) how much of the excess reactant(s) is needed
- excess reactant – the reactant left over after the reaction is complete
T.
1)
2
Fe(s) + 3 Cl2(g)  2 FeCl3(s)
If you begin with 5.63 grams of iron and 2.36 liters of chlorine gas:
a) What is the limiting reactant? What is the excess reactant?
2 FeCl 3
1 Fe
"
= 0.101 FeCl 3
55.8 g Fe
2 Fe
2 FeCl 3
1 Cl 2
2.36 L Cl 2 "
"
= 0.0702 FeCl 3 ! LR - Cl 2 , XR - Fe
22.4 L Cl 2
3 Cl 2
5.63 g Fe "
b) How many grams of iron(III) chloride do you expect?
0.0702 FeCl 3 !
162 g FeCl 3
= 11.4 g FeCl 3
1 FeCl 3
c) How many grams (liters if it’s a gas) of the excess reactant will be left over?
0.0702 FeCl 3 "
2 Fe
55.8 g Fe
"
= 3.92 g Fe used
2 FeCl 3
1 Fe
5.63 g Fe ! 3.92 g = 1.71 g Fe left
UNIT IV – STOICHIOMETRY
S.
1
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
If you begin with 3.77 moles of propane and 20.0 moles of oxygen:
a) What is the limiting reactant? What is the excess reactant?
3.77 C 3 H 8 !
20.0 O 2 !
3 CO 2
= 11.3 CO 2 " LR - C 3 H 8 , XR - O 2
1 C3H8
3 CO 2
= 12.0 CO 2
5 O2
b) How many liters of carbon dioxide do you expect? How many grams of water?
11.3 CO 2 !
22.4 L CO 2
= 253 L CO 2
1 CO 2
11.3 CO 2 !
4 H 2 O 18 g H 2 O
!
= 271 g H 2 O
3 CO 2
1 H 2O
c) How many liters of the excess reactant will be left over?
11.3 CO 2 !
5 O2
= 18.8 O 2
3 CO 2
20.0 O 2 " 18.8 O 2 = 1.2 O 2 !
22.4 L O 2
= 26.9 L O 2 left
1O2
6
UNIT IV – STOICHIOMETRY
7
VII. Percent Composition - tells you what the compound is made of
* resulting percents measure the percent composition of the compound by mass
* e.g. if a compound is composed of 55% copper, that means that a 100g sample of the
compound will contain 55g of copper atoms
S. Find the percent composition of the following - round your percentages to 3 sig figs
1) Al(NO3)3
Formula mass = 27.0 + 3 (14.0) + 9(16.0) = 213
27.0
x100 = 12.7% Al
213
3(14.0) 42
%N =
=
x100 = 19.7% N
213
213
9(16.0) 144
%O =
=
x100 = 67.6%O
213
213
% Al =
2) calcium phosphate
Ca3(PO4)2  3 (40.1) + 2(31) + 8(16) = 310
120
x100 = 38.8%Ca
310
62
%P =
x100 = 20.0% P
310
128
%O =
x100 = 41.3%O
310
%Ca =
3) Copper ore is primarily the mineral chalcopyrite (CuFeS2 ). Calculate the number of kilograms of
copper (Cu) contained in 3.71 x 103 kg of chalcopyrite.
Formula mass = 63.5 + 55.8 + 2(32.1) = 183.5
%Cu =
63.5
= 0.346 ! 3.71x103 kg = 1284 kg Cu
183.5
VIII. Laws of Chemical Combination
A) Law of Definite Proportions – different samples of the same compound will always contain
the same proportion of elements by mass
B) Law of Multiple Proportions – if two elements can combine to form more than one compound,
the masses of one element that combine with a fixed mass
of the other element are in the ratio of small whole numbers
* why is that?  MOLES
CO vs. CO2: both contain C & O, just in different ratios
IX. Experimental Determination of Atomic and Molecular Masses
* uses a mass spectrometer to separate atoms of different masses and determine the percent of the
sample composed of that type of atom
X. Finding Formulas of Compounds using laboratory data
* empirical formula - simplest whole number ratio of elements (not necessarily the actual
formula)
EXAMPLE: the empirical formula for C2H4 ( 2:4 ) is CH2 ( 1:2 )
S. What is the empirical formula for
1) acetylene (C2H2)
2) vitamin C (C6H8O6)
* simply reduce the ratio to its lowest possible value
CH
C 3 H4 O3
3) benzene C6H6
CH
UNIT IV – STOICHIOMETRY
8
* molecular formula - actual molecular makeup of the compound; will be a multiple
of an empirical formula
* List 3 possible molecular formulas for the empirical formula: C2H5
C2H5 , C4H10, C6H15, etc.
T.
1) If you have a compound which is composed of 76.1g of nitrogen and 173.9g of oxygen, what is its
empirical formula? If its formula mass is 92.0 g/mol, what is the molecular formula?
1 mol N
FM empirical = 14 + 2(16) = 46
= 5.44 mol N ÷ 5.44 = 1 N
14 g N
 NO2 FM molecular 92
=
= 2 ! multiply e.f. by 2
1 mol O
173.9 g O !
= 10.9 mol O ÷ 5.44 = 2 O
FM empirical 46
16 g O
76.1 g N !
N2 O4
S.
1) If you have a compound which is composed of 80.0% carbon and 20.0% hydrogen, what is its empirical
formula? If its formula mass is 30.0 g/mol, what is its molecular formula?
1 mol C
= 6.67 C ÷ 6.67 = 1 C
12 g C
 CH3
1 mol H
20.0 g H !
= 20 mol H ÷ 6.67 = 3 H
1g H
80.0 g C !
FM ef = 15
30
=2
15
mf ! C 2 H 6
2) If you have a compound which is composed of 58.0% sodium and 42.0% oxygen, what is its empirical
formula? If its formula mass is 78.0 g/mol, what is its molecular formula?
1 mol Na
= 2.52 Na ÷ 2.52 = 1 Na
23 g Na
 NaO
1 mol O
42.0 g O !
= 2.625 O ÷ 2.52 = 1 O
16 g O
58.0 g Na !
FM ef = 39
78
=2
39
mf ! Na 2 O 2
3) A compound composed of the elements carbon and hydrogen is known to be composed of 88.8% by
mass of carbon atoms and 11.2% by mass of hydrogen atoms.
The molar mass of this compound is known to be between 100 and 110 g. Determine its
molecular formula and the exact molar mass of the compound.
1 mol C
= 7.4 C ÷ 7.4 = 1 C x 2 = 2 C
12 g C
 C 2 H3
1 mol H
11.2 g H !
= 11.09 mol H ÷ 7.4 = 1.5 H x 2 = 3 H
1.01 g H
FM ef = 27
88.8 g C !
110
= 4.07 ! round to 4
27
mf ! C 8 H 12
UNIT IV – STOICHIOMETRY
XI. Concentration and Dilution of Solutions
A. Concentration of Solutions – most often, we express it as molarity (moles solute / liters sol’n)
1) What is the molarity of a solution in which you dissolve 10.0g of glucose (C6H12O6) in 50θ mL of
water?
10.0 g gluc !
1 mole gluc 0.0556 mol
=
= 0.111M
0.500 L
180 g gluc
B. Moles and Solutions
moles solute = molarity x volume (in liters)
1) How many moles of glucose are present in 500.0 mL of a 2.50 M solution?
n = MV = (2.50 mol )(0.500 L) = 1.25mol
L
* The above examples apply to nonelectrolytes, so what if an electrolyte dissociates in
water?
KCl(s)  K+ + Cl-
1 mol of KCl will dissociate to give 1 mol of K + and
1 mol of Cl- ions, so in 1 liter of solution:
+
[K ] = 1 M and [Cl-] = 1 M
Ba(OH)2(s)  Ba+2 + 2 OHIn 1 liter of solution:
[Ba+2] = 1 M and [OH-] = 2 M
C. Dilution of Solutions
MiVi = MfVf
1) Describe how you would prepare 50θ mL of a 1.75 M HNO3 solution beginning with an 8.61M stock
solution of HNO3.
(8.61M )V = (1.75M )(500ml )
(1.75M )(500ml )
V=
= 102ml stock soln
8.61M
500 mL - 102 mL = 398 mL water
S.
1) Concentrated HCl is a 12 M solution. If you need 50θ mL of 1.0 M solution for your experiment,
how much water and concentrated acid is required?
(12 M )V = (1.0 M )(500ml )
(1.0 M )(500ml )
V =
= 42ml stock soln
12 M
500 mL - 42 mL = 458 mL water
9
UNIT IV – STOICHIOMETRY
10
XII. Acid-Base Titrations
MaVa = MbVb
or moles acid = moles base
* standard solution – solution whose concentration is known
* unknown solution – solution whose concentration is determined by titrating with
the standard solution to the equivalence point
T.
1) If you titrate 50.0 mL of an unknown HCl solution with 34.2 mL of 3.00 M NaOH, what is the
concentration of the HCl?
M (50.0mL) = (3.00 M )(34.2mL)
M = 2.05M
2) In an experiment, a student finds that 23.5 mL of 1.00 M HCl is needed to neutralize some solid
Ba(OH)2. What mass of the Ba(OH)2 was used?
n = MV = (1.00 mol )(0.0235 L) = 0.0235mol H + ! will be neutralized by 0.0235 mol OH L
1 Ba(OH) 2 171.3g Ba(OH) 2
0.0235 mol OH - !
!
= 2.01 g Ba(OH) 2
1 Ba(OH) 2
2 OH S.
1) If you titrate 4θ drops of an unknown vinegar (HCH3COOH) solution with an average of 25.4
drops of a 0.600M standard NaOH solution, what is the concentration of the acid?
M (40dr ) = (0.600 M )(25.4dr )
M = 0.38M
2) Calculate the volume of 1.420 M NaOH needed to titrate 25.00 mL of 4.500 M H2SO4.
2 H+
1 OH n = MV = (4.500 mol )(0.02500 L) = 0.1125mol H 2 SO 4 !
!
= 0.225 mol OH +
L
1 H 2 SO 4 1 H
0.225 mol OH = 1.420 M
V
0.225 mol OH = V = 0.158 L
1.420 M
3) Potassium hydrogen phthalate (KHC8H4O4) is a white, soluble solid monoprotic acid often used to
standardize NaOH solutions. If 0.5468g of KHP is needed to completely neutralize 23.48 mL
of a NaOH solution. What is the concentration of the NaOH solution?
0.5468 g KHP !
1 KHP
1 OH - 2.68 x10 "3 mol OH !
=
= 0.1140 M
0.02348 L
204.2 g KHP 1 KHP
UNIT IV – STOICHIOMETRY
11
XIII. Redox Titrations – the equivalence point is determined when the oxidizing agent is completely
reduced
* MnO4- is good for this because it is deep purple, and when reduced to Mn+2, it shows
colorless to light pink
T.
1) A 16.42 mL volume of 0.1327 M KMnO4 solution is needed to oxidize 20.00 mL of a FeSO4 solution
in an acidic medium. What is the concentration of the FeSO4 solution?
The net ionic equation is 5 Fe+2 + MnO4- + 8 H +  Mn+2 + 5 Fe+3 + 4 H2O
5 Fe +2
!
!3
mol
n = MV = (0.1327
)(0.01642 L) = 2.18 x10 mol MnO 4 "
= 0.0109 mol Fe + 2
!
L
1 MnO
4
0.0109 mol Fe
0.02000 L
+2
= 0.545M
S.
1) Iron(II) can be oxidized by an acidic K2 Cr2O7 solution according to the net ionic equation:
Cr2O7-2 + 6 Fe+2 + 14 H+  2 Cr+3 + 6 Fe+3 + 7 H2O
If it takes 26.0 mL of 0.0250M potassium dichromate to titrate 25.0 mL of a solution containing
Fe+2, what is the molar concentration of Fe+2?
n = MV = (0.0250 mol )(0.0260 L) = 6.50 x10 ! 4 mol Cr2 O 7
L
!2
"
6 Fe +2
1 Cr2 O 7
!2
= 0.0039 mol Fe + 2
0.0039 mol Fe +2
= 0.156 M
0.0250 L
2) If you have 100.0 mL of a 0.100M potassium dichromate solution, how many grams of iron(II)
nitrate do you need to dissolve in 100.0 mL of water to titrate this solution?
n = MV = (0.100 mol )(0.1000 L) = 0.0100mol Cr2 O 7
L
0.0600 Fe + 2 !
!2
"
6 Fe +2
1 Cr2 O 7
!2
= 0.0600 mol Fe + 2
2 Fe(NO 3 ) 2 179.8 g Fe(NO 3 ) 2
!
= 10.79 g Fe(NO 3 ) 2
1 Fe(NO 3 ) 2
1 Fe + 2
3) A sample of iron ore weighing 0.2792g was dissolved in dilute acid solution, and all the iron(II) ions
were converted to iron(III) ions. The solution required 23.30 mL of 0.0194 M potassium
dichromate for the titration. What is the percent by mass of iron in the ore?
n = MV = (0.0194 mol )(0.02330 L) = 4.52 x10 ! 4 mol Cr2 O 7
L
0.00271 Fe + 2 !
!2
"
6 Fe +2
1 Cr2 O 7
!2
55.8 g Fe +2 0.1513 g Fe +2
=
x100% = 54.2%
0.2792g Ore
1 Fe + 2
= 0.00271 mol Fe + 2
UNIT IV – STOICHIOMETRY
12
XIV. Gravimetric Analysis – an analytical procedure that involves measurement of mass of a substance
quantitative analysis – determination of the amount or concentration of a substance in a sample
* by selectively precipitating a certain desirable ion, you can determine the composition:
T,
1) A sample of 0.5662g of an ionic compound containing chloride ions, but an unknown metal identity
is dissolved in water and treated with an excess of AgNO3. If the mass of the AgCl precipitate that
forms is 1.0882g, what is the percent by mass of Cl in the original compound?
Ag+ + Cl-  AgCl(s)
(1.0882 g AgCl)(0.247) =
% Cl in AgCl :
35.45 g Ag
= 0.247
143.4 g AgCl
0.2690 g Cl
= 47.5%Cl
0.5662 g XCl
2) The concentration of lead ions (Pb+2) in a sample of polluted water that also contains nitrate ions (NO3-)
is determined by adding solid sodium sulfate (Na2SO4) to exactly 50θ mL of the water.
a) Write the net ionic equation for the reaction.
Pb+2 + SO4-2  PbSO4
b) Calculate the molar concentration of Pb+2 if 0.450g of Na2SO4 was needed for the complete
precipitation of Pb+2 ions as PbSO4
"2
+2
1 Na 2 SO 4
1 SO 4
1 Pb +2
0.450 g Na 2 SO 4 !
!
!
= 0.00317 mol Pb
= 0.00633M
"2
0.500 L
142.1 g Na 2 SO 4 1 Na 2 SO 4 1 SO 4
3) If 30.0 mL of 0.150 M CaCl2 is added to 15.0 mL of 0.100 M AgNO3, what is the mass of AgCl
precipitate formed?
Net ionic equation:
Ag+ + Cl-  AgCl
nCl "
2 mol Cl = MV = (0.0300ml )(0.150 M ) = 0.00450 mol CaCl 2 !
= 0.00900 mol Cl 1 mol CaCl 2
nCl "
1 mol Ag +
= MV = (0.0150ml )(0.100 M ) = 0.00150 mol AgNO 3 !
= 0.00150 mol Ag +
1 mol AgNO 3
* Because the Ag+/Cl- ratio in the above equation is 1:1, and the Ag+ is smaller, Ag + is the LR, and
therefore determines the amount of precipitate formed:
0.00150 Ag + !
1 AgCl 143.45g AgCl
!
= 0.215 g AgCl
1 AgCl
1 Ag +
S
1) How many grams of NaCl are required to precipitate most of the Ag+ ions from 2.50 x 102 mL of
0.0113 M AgNO3 solution? Write the net ionic equation for the reaction.
1 Cl " 1 NaCl 58.45g NaCl
n = MV = (0.0113 mol )(0.250 L) = 0.002825 Ag + !
!
!
= 0.165g NaCl
L
1 NaCl
1 Ag + 1 Cl "
2) A sample of 0.3220g of an ionic compound containing the bromide ion is dissolved in water and
treated with an excess of AgNO3. If the mass of the AgBr precipitate formed is 0.6964g, what
is the percent by mass of Br in the original compound?
1 AgBr
1 Ag + 1 Br - 79.9 g Br - 0.2961 g Br 0.6964 g AgBr !
!
!
!
=
x100% = 92.0%
0.3220 g XBr
187.9 g AgBr 1 AgBr 1 Ag +
1 Br -