Download KCl + O KClO 3 → However, this equation is not balanced, since

Assignment 50
CHEMICAL EQUATIONS, AND CALCULATIONS FROM THEM
The chemical properties of a substance describe how the substance becomes changed under various circumstances. The
changes which occur are the chemical reactions the substance enters into, and these chemical reactions are represented
in chemical shorthand notation as chemical equations. In this shorthand notation, the formulas of the substances are
used to represent all substances involved in the process.
It is conventional in the equation to place the substances mixed (the reactants) on the left and the substances formed
(the products) on the right. At the present time, a sign, = or ➙, will be used between reactants and products, and this
will be read as "produces" or "yields" or "forms". It will be shown later that it is incorrect to imply complete conversion
of reactants into products in many chemical reactions, but for the present, only reactions which do proceed completely
will be considered in numerical problems.
Writing a balanced chemical equation:
In order to write a chemical equation, it is essential to know what substances are reacting and what substances are
formed. It is only through experiment that a knowledge of what occurs can be discovered ─ either your experiment or
the experiments of others (which can be found in textbooks). As more experience becomes gained and as more of the
theoretical basis of the science of chemistry is presented, you will be able to grasp more readily what might occur in the
chemical reaction, but definite proof must still be sought in experiment. In the initial work with equations, the formulas
of all substances entering into the chemical reaction will be given.
All chemical equations must adhere to the natural law of conservation of mass. In terms of the fundamental units of
matter (i.e., the atoms of the substances entering the chemical reaction), this law means that each atom placed in the
chemical reaction as a reactant must appear in the products of the reaction. Atoms are neither created nor destroyed in
ordinary chemical reactions, but the various atoms must be rearranged into new compounds in order to have any
chemical reaction at all. The process of adjusting the numbers of each kind of atom on both sides of the equation is
called balancing the equation.
As an example, a possibly familiar reaction in the laboratory is that of heating potassium chlorate (KClO3) to produce
oxygen (O2) and potassium chloride (KCl). This statement of the chemical reaction has been proved by experiment.
This reaction may be represented by the equation:
KClO3 → O2 + KCl
However, this equation is not balanced, since there are three atoms of O on the left and only two on the right, while the
K and Cl atoms are balanced.
The balanced equation for the reaction is :
2 KClO3 → 3 O 2 + 2 KCl
In this equation there are 2 K atoms, 2 Cl atoms, and 6 O atoms on both sides of the equation; and this equality, for each
kind of atom present, is the criterion of a balanced equation if no charged particles appear in the statement of the
equation.
Chemical equations are read in terms of the substances whose formulas are shown. Just as the formula may stand for a
single discrete unit of the substance or for one mole of the substance, the equation may be read either in single discrete
units or in moles. The equation above is read:
1)
2 "molecules" (or formula groupings) of potassium chlorate yield 3 molecules of oxygen and 2
"molecules" (or formula groupings) of potassium chloride.
OR
2)
2 moles of potassium chlorate produce 3 moles of oxygen and 2 moles of potassium chloride.
This second method ─ reading the equation in moles ─ will be more useful at the present time.
01-111-50-1
To achieve a balanced equation, it is necessary to know (or be given as information) not only the formulas of all
substances in the chemical reaction but also how much of the reacting substances are required for a balancing of the
reactant and product formulas. As the course develops, logical reasons for the occurrence of chemical reactions and
methods of figuring out what happens will be given. But until this development takes place, the balancing of a chemical
equation must be done by trial-and-error from a skeleton equation, which will show the formulas of every substance
and (where necessary) the number of moles of reactants entering the reaction. In the balancing process, each formula
must remain intact without alternation; only the coefficient to the left of the formula may be altered.
Examples:
Balance each skeleton equation below (i.e., fill in the blanks with appropriate coefficients) and read each balanced
equation.
1)
Given:
Balance:
__H2 + __O2 → __H2O
2 H on left and 2 H on right are balanced, but the 2 O on left and 1 O on right are not. Multiply H2O by
2 to get 4 H atoms and 2 O atoms, which then balances the 2 O atoms on the left. But the H atoms now
are out-of-balance.
__H2 + _1_O2 → _ 2_H2O
Multiply H2 by 2 ; this gives the necessary balance of 4 H and 2 O atoms on each side. The balanced
equation is
_ 2_H2 + __O2 → _ 2_H2O
2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
Read:
2)
Given: H3 PO3 → H3 PO 4 + PH 3
Balance:
Note first the O atoms, which occur in only one reactant, H3PO3, and must appear in only one product,
H3PO4. Multiply H3PO3 by 4 , giving (4 3) = 12 O atoms in the reactant, H3PO3. All 12 O atoms
must appear in the H3PO4 and must produce 3 H3PO4 (12 O atoms).
Now balance the P atoms without altering the oxygen balance. The P atoms on the left (4 of them, in
the 4 H3PO3) go into H3PO4 (3 of them, since there must be 3 H3PO4) and into PH3 (the 1
remaining P). Thus, the balancing requires 1 PH3 .
4 H3 PO3 → 3 H3 PO4 + 1 PH3
The three coefficients have been established; the H atoms are seen to be in balance now also. On the left
are (4 3) = 12 H atoms. On the right there are, in total, also 12 H atoms ─ (3 3) = 9 H atoms from
the 3 H3PO4 and (1 3) = 3 H atoms from the 1 PH3 .
So, the balanced equation as normally written is:
4 H3 PO3 → 3 H 3 PO4 + PH3
4 moles of H3PO3 (phosphorous acid) form 3 moles of H3PO4 (phosphoric acid) and 1 mole of PH3
(phosphine).
Read:
3)
Given: 2 C3 H 6 + O2 →
CO2 + H 2 O
Balance:
6 C on left, 6 CO2 formed
12 H on left, 6 H2O formed
On the right there are now a total of 18 O atoms; these include (6 2) = 12 in the CO2 plus (6 1) =
6 in the H2O. To complete the balance, there must be a total of 18 O atoms on the left ─ therefore, 9 O2
are needed. The balanced equation is: 2 C3 H 6 + 9 O 2 → 6 CO 2 + 6 H 2 O
Read:
2 moles of C3H6 (propene) react with 9 moles of oxygen to produce 6 moles of carbon dioxide and 6
moles of water.
01-111-50-2
4)
Given:
C3 H 6 + O 2 →
CO 2 + H 2 O
(This is similar to example (3) except with no coefficient given.)
Balance:
If 1 C3H6 is assumed for a start, then 3 C on left form 3 CO2 on right, and 6 H on left form 3 H2O.
But this yields an odd number of oxygen atoms on the right, unsuitable for balancing with O2 on the left.
Therefore, assume 2 C3H6 , instead. The balancing then proceeds as in example (3).
From ionic reactions, only the essential ions are put down in the chemical equations; however, it must be remembered
that, in order to work with an ion, some other ion must be present so as to counterbalance the ionic charge, although this
other ion takes no part in the reaction. Ions which do not react are called spectator ions.
An additional criterion for the balance of equations involving charged ions is the conservation of charge. The total
net charge (algebraic sum) must be the same for the reactants and for the products of the reaction.
Examples:
+
1)
Given: Ag + Cl- → AgCl (s)
Balanced as written. Net charge on reactants is zero (since 1+ and 1− together have a net charge of zero). Net
charge on product is also zero.
+
Read: 1 mole of Ag (silver ion) reacts with 1 mole of Cl! (chloride ion) to produce 1 mole of solid AgCl
(silver chloride).
Note: The reagent solutions used in this reaction might be silver nitrate (Ag NO3−) and sodium chloride
+
+
+
(Na Cl−). Then the NO3− (ions needed to counterbalance Ag in the reagent) and the Na (ions needed
−
to counterbalance Cl in the reagent) would be considered spectator ions (for they do not react during
the mixing).
+
2)
Given:
5 Fe2+ + MnO-4 + 8 H + → 5 Fe3+ + Mn 2+ + 4 H 2 O
Balanced as written. Net charge on reactants is (5 x 2+) + (1 x 1−) + (8 x 1+) = 17+, and net charge on product
is (5 x 3+) + (1 x 2+) = 17+ .
2+
+
Read: 5 moles of Fe (ferrous ion) reacts with 1 mole of MnO4− (permanganate ion) and 8 moles of H
+
3+
2+
(hydronium ion, H3O , in solution) to produce 5 moles of Fe (ferric ion), 1 mole of Mn (manganous ion),
and 4 moles of water.
Calculations from balanced equations:
A chemical equation gives a large amount of information about a chemical reaction. Not only does it show what
substances are involved in the reaction, but, when balanced, it also shows how many moles of each substance takes part.
The balanced equation, therefore, gives the proportions (in moles) among the various reactants and products in the
chemical reaction. These proportions are exacts in terms of moles, but are relatives in terms of the actual amounts
entering the reaction. As soon as the amount of any one substance is fixed, the mole proportions of all other substances
in the reaction become fixed. Each of these proportions may be used as a conversion factor to determine the amounts
of the other substances reacting with or produced from the fixed (or given) substance. Thus, a balanced equation can be
used to predict the amount of material which is produced from the amount of material reacted. Or, from knowing the
amount of product desired, the amount of reactant needed can be determined.
In the examples and problems of this assignment, it will be assumed that the chemical reactions will proceed completely
to form the products shown and that enough materials are present to achieve the complete conversion of the amount of
the given substance according to the balanced equation. Such calculations are often referred to as theoretical yield
calculations.
01-111-50-3
Examples:
Given this reaction: C6 H12 O6 → 2 C2 H5 OH + 2 CO 2
This equation indicates that, for each mole of sugar (C6H12O6) used in this reaction, two moles of ethyl alcohol
(C2H5OH) and two moles of carbon dioxide are formed. By means of this relationship, if one knows the amount
of any one of the substances involved, the amount of any of the other materials can be calculated.
1)
How many moles of C2H5OH will be produced from 3 moles of C6H12O6?
From the balanced equation, one obtains the relationship:
1 mol of C6H12O6 yields 2 mol C2H5OH.
This relationship may be expressed as a conversion factor:
 2 mol C2 H5 OH 


 1 mol C6 H12 O6 
or, equally true when inverted,
 1 mol C6 H12 O6 


 2 mol C2 H5 OH 
Dimensional analysis shows that the desired units ─ moles of C2H5OH ─ can be obtained by multiplication in the
following manner:
 2 mol C2 H5 OH 

Number of moles C2 H5 OH = number of moles C6 H12 O6 x 
 1 mol C6 H12 O6 
Therefore, to solve the question, one writes:
 2 mol C2 H5 OH 
 = 6 mol C2 H5 OH
? moles C2 H5 OH = 3 mol C6 H12 O6 x 
 1 mol C6 H12 O6 
2)
How many moles of C6H12O6 would be needed to produce 0.50 moles of CO2?
From the balanced equation, the conversion factor relating C6H12O6 and CO2 may be written:
 1 mol C6 H12 O6 


 2 mol CO 2 
or, equally true when inverted,
 2 mol CO 2 


 1 mol C6 H12 O6 
Note: The desired conversion factor is always the number of moles of what is to be solved for divided by the
number of moles of what is given.
Thus, to answer the question, one writes:
 1 mol C6 H12 O6 
? moles C6 H12 O6 = 0.50 mol CO2 x 
 = 0.25 mol C6 H12 O6
 2 mol CO2 
The numerals in a conversion factor are exact numbers and do not determine the number of significant
figures in a final answer.
3)
How many moles of CO2 are produced if 3.5 moles of C2H5OH are produced?
From the balanced equation, the desired conversion factor is:
 2 mol CO 2 


 2 mol C2 H5 OH 
 2 mol CO2 
? moles CO 2 = 3.5 mol C2 H5 OH x 
 = 3.5 mol C2 H5 OH
 2 mol C2 H5 OH 
01-111-50-4
Since a chemical balance does not weigh in mole units, but rather in grams, it frequently becomes necessary to make
conversions between numbers of moles and grams of weight.
Examples:
Given the reaction:
2 H3 AsO3 + 3 H 2 S → 6 H2 O + As2 S3
2 moles + 3 moles → 6 moles + 1 mole
The equation indicates:
From this mole relationship, the weight of the substances involved can be calculated.
1)
How many grams of H2O can be produced by the reaction of 0.10 mole of H3AsO3?
From the given data (moles of H3AsO4), find first the number of moles of H2O;
 6 mol H 2 O 

 1 is obtained from the balanced equation.
 2 mol H3 AsO3 
 6 mol H 2 O 
Number of moles H2 O = given number of moles H3 AsO3 x 

 2 mol H3 AsO3 
the conversion factor
 6 mol H 2 O 
= 0.10 mol H3 AsO3 x 

 2 mol H3 AsO3 
 0.10 x 6
( which means you have, thus far, 
2


 moles of H 2 O )

Then, find the weight of H2O by knowing its molecular weight is 18.0 g/mol H2O:
 18.0 g H2 O 
Number of grams of H 2 O = number of moles of H 2 O x 

 1 mol H2 O 
The entire problem can be, and should be, set-up in a single line set-up, thus:
 6 mol H 2 O   18.0 g H 2 O 
? grams H 2 O = 0.10 mol H 2 AsO3 x 
 x 
 = 5.4 g H 2 O
 2 mol H3 AsO3   1 mol H 2 O 
(Observe that all the significant information is present ─ what you are solving for, what is known data,
what conversions are being made, and the final result.) The consecutive factors are arranged into
place, one after the other, step-by-step. If the factors are written correctly, all units should cancel except
the final units desired. Then the final value is solved numerically by a single computation at the
conclusion only ─ arithmetically a simple task:
 0.10 x 6 x 18.0 
 = 5.4

2 x1


(two sig . figs . because of the 0.10 value)
01-111-50-5
2)
How many moles of H3AsO3 are needed to react completely with 10.0 grams of H2S?
From the given weight of H2S , find first the number of moles of H2S , by knowing the MW = 34.0 g/mol H2S :
 1 mol H 2 S 

Number of moles of H 2 S = 10.0 grams of H 2 S x 
 34.0 g H 2 S 
Then, find the number of moles H3AsO3 which react with this number of moles H2S ,
by using the conversion factor from the balanced equation:
 2 mol H3 AsO3 


 3 mol H2 S 
A single-line set-up is written in order to show all the key features:
 1 mol H 2 S   2 mol H3 AsO3 
 x 
? moles H3 AsO3 = 10.0 g H 2 S x 
 = 0.196 mol H3 As O3
 34.0 g H 2 S   3 mol H 2 S 
Note that the dimensions cancel properly to yield the desired units, and the sequence of values shown in
the set-up is easy to solve:
 10.0 x 1 x 2 
 = 0.196

 34.0 x 3 
(three sig . figs . because of 10.0 value)
An alternate set-up is also satisfactory but is less immediately evident to the eye:
? mol H3 AsO3 =
3)
10.0 g H2 S
 2 mol H3 AsO3 
x
 = 0.196 mol H3 AsO3
( 34.0 g H 2 S / mol H 2 S )  3 mol H 2 S 
How many grams of As2S3 can be produced from 30.0 grams of H3AsO3?
First, find the number of moles of H3AsO3 (MW of H3AsO3 = 126 g/mol).
 1 mol H3 AsO3 

Number of moles H3 AsO3 = number of grams of H3 AsO3 x 
126
g
H
AsO
3
3


Then, find the number of moles of As2S3 from the number of moles of H3AsO3 ;
 1 mol As2 S3 

 is obtained from the balanced equation.
 2 mol H3 AsO3 
 1 mol As2 S3 
Number of moles As2 S3 = number of moles H3 AsO3 x 

 2 mol H3 AsO3 
the conversion factor
Finally, find number of grams of As2S3 (MW of As2S3 = 246 g/mol).
 246 g As2 S3 
Number of grams of As2 S3 = number of moles of As2 S3 x 

 1 mol As2 S3 
The single-line set-up shows all these conversions and is solved as a single calculation:
 1 mol H3 AsO3   1 mol As2 S3   246 As2 S3 
 x 
? g As2 S3 = 30.0 g H3 AsO3 x 
 x 
 = 29.3 g As2 S3
 126 g H3 AsO3   2 mol H3 AsO3   1 mol As2 S3 
or:
? g As2 S3 =
 1 mol As2 S3 
30.0 g H3 As O3
x 
 x (246 g/mol As2 S3) = 29.3 g As2 S3
(126 g/mol H3 AsO3)  2 mol H3 AsO3 
01-111-50-6
Note, in the overall set-ups above, that each dimension is actually a "double dimension" -- a unit and a formula.
Each conversion factor alters either the unit or the formula; but the unit and formula are never both altered at the same
instant. In chemical calculations, whenever one wishes to interrelate two formulas, the accompanying unit must be
mole. Therefore, the sequencing of the conversion factors is important: conversion to the mol unit while retaining the
original formula, then conversion to the desired formula, then conversion to the final unit with that new formula. The
mol/mol conversions typically occur at the middle steps of the problem.
If the reaction involves gases, the gram molecular volume (GMV = 22.4 LSTP/mol of gas) may be used to convert
between numbers of moles and volume of gas under STP conditions, assuming that the gas behaves like an ideal gas.
Examples:
Given the reaction:
C2H5OH (l) +
3 O2 (g)
➙
3 H2O (Ρ) +
The equation indicates:
1 mole +
(liquid)
3 moles
(gas)
➙
3 moles
(liquid)
2 CO2 (g)
+
2 moles
(gas)
Note: The symbols, (g), (l), or (s), shown after the formula of a compound, stand for the state of the substance:
gas, liquid, or solid. (The symbol for liquid is often omitted.)
From these mole relationships, the volume relationships of the gases can be determined.
1)
How many moles of C2H5OH are needed to produce 6.0 liters of CO2 at STP? (see eqn above)
First, from the given liters of gas at STP, find moles of CO2 by using the GMV:
 1 mol CO 2 
Number of moles of CO2 = number of liters(STP) of CO2 x 

 22.4 LSTP CO2 
Then, find the number of moles C2H5OH necessary to produce this quantity of CO2 ; the reaction equation
 1 mol C2 H5 OH 

 2 mol CO2 
provides the mol/mol conversion factor: 
 1 mol C2 H5 OH 
Number of moles of C2 H5 OH = number of moles of CO 2 x 

 2 mol CO2 
The single-line set-up and calculation (with all volumes at STP) is written thus:
 1 mol CO2   1 mol C2 H5 OH 
? moles C2 H5 OH = 6.0 LSTP CO 2 x 
 x 
 = 0.14 mol C2 H5 OH
 22.4 LSTP   2 mol CO2 
or :
? moles C2 H5 OH =
 1 mol C2 H5 OH 
6.0 LSTP CO2
x 
 = 0.14 mol C2 H5 OH
( 22.4 LSTP /mol CO2 )  2 mol CO2 
01-111-50-7
2)
How many liters of O2 at STP will be needed to produce 0.30 moles of H2O?
From the given moles of H2O, find the moles of O2 needed (the conversion factor is provided by the reaction
equation):
 3 mol O2 
Number of moles of O2 = number of moles of H2 O x 

 3 mol H2 O 
Then, find liters(STP) of O2, by using the GMV relationship:
 22.4 LSTP O2 
Number of liters(STP) of O2 = moles O2 x 

 1 mol O2 
The single-line answer is:
 3 mol O2   22.4 LSTP O2 
? liters(STP) O2 = 0.30 mol H 2 O x 
 = 6.7 LSTP O2
 x 
 3 mol H2 O   1 mol O2 
3)
How many grams of H2O are produced when 7.0 liters of CO2 are produced at STP?
First, from the given liters(STP) of CO2 , find the moles of CO2 by using the GMV.
Then, convert the formula to find moles of H2O by using the appropriate mol/mol conversion from the balanced
equation.
Finally, from the moles of H2O, find the grams of H2O by using the molecular weight of H2O.
The single-line answer with all these conversions is written thus:
 1 mol CO2   3 mol H2 O   18.0 g H 2 O 
? grams H 2 O = 7.0 LSTP CO 2 x 
 x 
 x 
 = 8.3 g H2 O
 22.4 LSTP CO2   2 mol CO 2   1 mol H 2 O 
4)
How many liters O2 at STP would be required, with excess C2H5OH , to form 4.0 liters CO2 at STP?
First, from the given liters(STP) of CO2 , find the moles of CO2 by using the GMV.
Then, convert the formula to find moles of O2 by using the appropriate mol/mol conversion from the balanced
equation.
Finally, from the moles of O2 , find the liters(STP) of O2 by using the GMV again.
The single-line answer showing all these conversions is written thus:
 1 mol CO2   3 mol O2   22.4 LSTP O2 
? LSTP O2 = 4.0 LSTP CO 2 x 
 = 6.0 LSTP CO 2
 x 
 x 
 22.4 LSTP CO 2   2 mol CO 2   1 mol O2 
Note how easily the number 22.4 cancels in this single-line form.
Calculations with reactions in a sequence:
Often the desired product in a laboratory experiment is obtained by means of a series of reactions, when several
reaction equations may be involved. The mol/mol conversion factors may be determined by either of two methods:
1)
or
2)
use a sequence of mol/mol conversion factors ─ one from each equation in the series;
on paper, combine the chemical equations so as to create a single overall equation for the reaction; then
find the overall conversion factor from that one equation.
(Important note: Any conversion factor must involve only a single, balanced, chemical equation.)
01-111-50-8
Example:
Given this three-step reaction sequence:
a)
Mg2C3
+
4 H2O
➙
C3H4 +
2 Mg(OH)2
b)
2 C3H4
+
5 O2
➙
6 CO +
4 H2O
c)
2 CO
+
O2
➙
2 CO2
For these equations, some single-equation conversion factors might be:
from (a):
from (b):
 2 mol Mg(OH )2 




 1 mol Mg 2 C3 
 6 mol CO 


 2 mol C3 H 4 
or
 1 mol C3 H 4 


 4 mol H 2 O 
 5 mol O2 


 4 mol H2 O 
etc.
etc.
 2 mol CO2 


etc.
 1 mol O2 
 mol CO2 
 , because
None of the equations provides a direct conversion factor for 

mol
Mg
C
3
2


from (c):
 2 mol CO 


 2 mol CO2 
or
or
CO2 appears only in equation (c), and Mg2C3 only in equation (a). No equation includes both formulas.
In order to calculate moles of CO2 from a given number of moles of Mg2C3 by means of the sequence of reaction
equations above, it is necessary to proceed as follows:
By method (1), devise a sequence of conversion factors, as a substitute for a direct conversion factor:
 ? mol CO 2

 ? mol Mg C
2 3

  1 mol C3 H 4   6 mol CO   2 mol CO 2 
 
 = 

  1 mol Mg C  x  2 mol C H  x  2 mol CO 
3 4
2 3 
 
from (a)
from (b)
from (c)
(Note that the dimensional units cancel appropriately and give a net relationship of
3 moles CO2 per 1 mole Mg2C3.)
OR
By method (2), construct a single overall equation, by canceling completely any substance which is formed in
one reaction and used up in a subsequent reaction. Thus, to obtain CO2 from Mg2C3 , the C3H4 must be
canceled from equations (a) and (b), and the CO must be canceled from equations (b) and (c).
Since equation (b) includes 2 C3H4 while equation (a) has only 1 C3H4 , multiply all of equation (a) by the exact
number, 2. Similarly, in order to cancel the 6 CO in equation (b), multiply all of equation (c) by 3 to give 3 x 2
CO = 6 CO there. (Note: On your paper, it is important to state what multiplications you are making!)
2 x (a) :
2 Mg2C3
+
8 H2O
➙
2 C3H4
+
4 Mg(OH)2
1 x (b) :
2 C3H4
+
5 O2
➙
6 CO
+
4 H2O
3 x (c) :
6 CO
+
3 O2
➙
6 CO2
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Adding these equations, and then canceling substances that appear on both sides of the arrow, gives this overall
equation:
2 Mg2C3
+
4 H2O
+
8 O2 (g)
➙
6 CO2
From this one overall equation comes the desired direct conversion factor:
+
4 Mg(OH)2
 6 mol CO2 




 2 mol Mg 2 C3 
By either method, (1) or (2), the conversion will produce the same computational effect. Each method gives this
resulting relationship: 3 moles CO2 per 1 mol Mg2C3 .
In the example above, all the given equations were multiplied by positive numbers. Note, however, it is permissible
also to multiply equations by negative numbers, which has the effect of reversing the reaction.
Example:
Burning of hydrogen to form water:
2 H2(g) + O2(g) ➙ 2 H2O(l)
Multiplying the equation by −1 gives the reverse reaction (as in the decomposition of water):
2 H2O(l) ➙
2 H2(g) + O2(g)
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