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>>
View the other Key Equations and Concepts in this
chapter
Chemical Equations
>> Parts of this equation/concept include:
| A. Balancing Chemical Equations |
A balanced chemical equation has the same number of each kind of
element on each side of the equation. Balancing is done by adding
the appropriate stoichiometric coefficients in front of the formulas.
The stoichiometric coefficients apply to the entire molecule and
should be whole numbers. Changing subscripts is not permitted because
that changes the identity of the compound.
Although there are systematic methods for this, they are normally
more difficult than trial and error. However, a few tips sometimes
prove useful.
- Start with the most complex compound and work to the simplest
compound.
- If a fraction will balance the reaction, use one. Multiply by
the denominator to get whole numbers.
- Treat polyatomic atoms as one group rather than as separate
elements.
- Water tends to react as H and OH, so balance in those parts.
- Ammonium ions can react as H and NH3, so watch for
those parts.
>> Example 1
Balance the following chemical equations.
- K2CO3 + Al(ClO4)3
 KClO4 + Al2(CO3)3
- Zn + HCl ZnCl2
+ H2
- C8H18 + O2
CO2 + H2O
- HF + Mg(OH)2
MgF2 + H2O
- NH4Cl + Ba(OH)2
NH3 + H2O + BaCl2
Solution:
-
Treat the polyatomic ions like one group. Note that the most
complex one is Al2(CO3)3,
(It has the most subscripts.)
To balance the aluminums, put a 2 in front of aluminum perchlorate.
To balance the carbonates, put a 3 in front of potassium
carbonate.
That means there are 6 potassiums, so a 6 in front of potassium
perchlorate is needed.
That balances it.
3 K2CO3 + 2 Al(ClO4)3
6 KClO4 + Al2(CO3)3
- You need two chlorines on the left, so a 2 in front of HCl will do it.
Zn + 2 HCl ZnCl2
+ H2
-
Start with C8H18 as the most complex
and O2 is the simplest.
To balance the carbons, put an 8 in front of the CO2.
To balance the hydrogens, put a 9 in front of the water.
That makes 16 + 9 = 25 oxygens on
the right. Since it is O2 on the left, 25/2 would
balance the number of oxygens. So do that as a temporary step.
C8H18 + 25/2 O2
8 CO2 + 9 H2O
Multiply each coefficient by 2.
2 C8H18 + 25 O2
16 CO2 + 18 H2O
-
The groups in this equation are H, F, Mg, and OH. It is easier
to see the relationships, by noting that water acts like HOH.
HF + Mg(OH)2
MgF2 + H–OH
So the OH can be balanced with 2 waters, and the fluorine
with 2 HF. That does it.
2 HF + Mg(OH)2
MgF2 + 2 H2O
-
Treat ammonium ion as NH3-H and water as H–OH.
NH3–HCl + Ba(OH)2
NH3 + H–OH + BaCl2
Balance the chlorine with 2 NH4Cl.
Balance the OH with 2 waters.
Balance the ammonia with 2 NH3.
That works.
2 NH4Cl + Ba(OH)2
2 NH3 + 2 H2O + BaCl2
>> back
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| B. Stoichiometric Calculations |
Stoichiometric calculations relate quantities of different compounds
to each other. Balanced chemical equations can be either ratios
of molecules or moles. It is helpful to keep track of both the unit
and the substance it refers to, and to change only one of those
at a time.
The following procedure is useful for doing stochiometric calculations
with dimensional analysis. A step in the procedure may be skipped
if it is not necessary. (For example, you don't have to convert
a unit to moles if the unit is moles!) It may also take more than
one conversion to accomplish some steps. (For example, before you
convert grams to moles, you may have to convert milligrams to grams.)
>> General Procedure
for Solving Stoichiometric Problems
- Balance the reaction.
- Convert starting unit to moles.
- Use molar mass to change grams to moles.
- Use Avogadro's number to change atoms or molecules to moles.
- Convert moles of starting substance to moles of substance you want.
- Use stoichiometric coefficients of balanced reaction.
- Use chemical formula for an element to compound ratio.
- Convert moles to unit you want.
- Use methods in step 2.
>> Example 2
How many grams of aluminum chloride can be made from 0.149 g
of hydrochloric acid?
Al(OH)3 + HCl
H2O + AlCl3
Solution:
Step 1. Balance the reaction.
Al(OH)3 + 3 HCl
2 H2O + AlCl3
Step 2. Convert the starting unit to moles.
The starting point is 0.149 g HCl.
The method to convert grams to moles is molar mass.
Since the substance is HCl, the molar mass of HCl is required.
The molar mass of HCl = 1.008 + 35.453 = 36.461 g/mol,
so 36.461 g = 1 mol HCl.
| 0.149 g HCl |
 |
|
Step 3. Convert the starting substance to the substance desired.
Started with HCl, the question asks about AlCl3.
The chemical reaction says that 3 moles HCl makes 1 mole AlCl3.
| 0.149 g HCl |
|
|
|
|
Step 4. Convert moles to units desired.
The question asks for grams, so molar mass is used.
Since the substance is now AlCl3, the molar mass
of AlCl3 is needed. The molar mass of AlCl3 = 26.98 + 3(35.453) = 133.34 g/mol.
| 0.149 g HCl |
|
|
|
|
|
|
= |
0.182 g AlCl3 |
The three significant figures were determined by the value from
the problem. The molar masses have more significant figures and
the mole-to-mole ratio has infinite significant figures.
>> Example 3
How many milligrams of carbon tetrachloride are made from 0.039
mole of Cl2?
Cl2 + CO2 CCl4 + O2
Solution:
Step 1. Balance the reaction.
2 Cl2 + C CCl4
+ O2
Step 2. Convert the unit given to moles.
The starting point is 0.039 mol Cl2
The unit is already moles, so move on to the next step.
Step 3. Convert the substance.
The balanced reaction says 2 moles Cl2 for one mole of CCl4.
| 0.039 mol Cl2 |
|
|
Step 4. Convert the units from moles.
The requested unit is milligrams, a variation on grams.
To convert moles to grams, molar mass is used.
The substance is now CCl4; its molar mass = 12.01 + 4(35.453) = 153.82 g/mol.
To convert grams to milligrams, use the SI relationship 1000
mg = 1 g.
| 0.039 mol Cl2 |
|
|
|
|
|
|
= |
3.0 x 103 mg CCl4 |
>> Example 4
How many micrograms of CaO are made when 3.6 x 1016
molecules of O2 are produced?
CaCO3 CaO + O2
Solution:
Step 1. Balance the equation.
CaCO3 CaO + O2
Sometimes it is balanced as it stands. Check anyway.
Step 2. Convert given units to moles.
The starting unit is molecules of O2.
Molecules are converted to moles with Avogadro's number,1 mol = 6.022 x 1023 molecules.
| 3.6 x 1016 molec O2 |
|
| 1 mol O2 |
|
| 6.022 X 1023 molec |
|
Step 3. Change the substance.
The balanced equation says 1 mol CaO to 1 mol O2.
| 3.6 x 1016 molec O2 |
|
| 1 mol O2 |
|
| 6.022 X 1023 molec |
|
|
|
Step 4. Convert to desired units.
Desired units are micrograms, a variation of grams—1 gram = 1,000,000  g.
Get to grams with molar mass. The substance is now CaO, so
40.08 + 16.00 = 56.08 g/mol.
| 3.6 x 1016 molec O2 |
|
| 1 mol O2 |
|
| 6.022 x 1023 molec |
|
|
|
|
|
|
| 106 g |
|
| 1 g |
|
= |
3.4 g CaO |
>> Determining Limiting
Reactants
Most reactions require more than one reactant. The amount of product
will be determined by which reactant is used up first. The excess
of the other reactant will not affect the amount of product.
Therefore to determine the amount of product, you must first determine
the reactant that is used up first, the limiting reactant. The procedure
for determining the amount of product is the same; it is just that
the starting point in step 2 must be the limiting reactant. There
are two ways to determine the limiting reactant.
>>> Method 1
In method 1, you compare the two reactants. Pick one (it doesn't
matter which!), and assume that this reactant will be all used
up. Then use the procedure for solving stoichiometric problems
to determine the amount of other reactant needed. (Do the calculation
to get the same units as those given in the problem.) Compare
your answer to the units given. If the answer is less than the
amount given in the problem, the reactant you started with is
the limiting reactant. If the answer is greater than the amount
given in the problem, the reactant you ended up with is the limiting
reactant. Notice that either way you determine the limiting reactant,
so it doesn't matter which you pick.
>> Example 5
What is the limiting reactant when 0.549 g potassium metal is
mixed with 0.669 g of Cl2 to make KCl?
K + Cl2 KCl
Solution:
Step 1. Balance the equation.
2 K + Cl2 2
KCl
Steps 2-4. Starting with grams K, going to grams Cl2,
so need molar masses of K = 39.10 g/mol and Cl2 = 2(35.453) = 70.906 g/mol.
Since stoichiometry says 0.498 g Cl2 are needed
and 0.669 g Cl2 are given in the problem (we need
less than we have), then K is the limiting reactant.
Steps 2-4. Starting with grams Cl2, going to grams
K.
Since stoichiometry says we need 0.738 g K and we only have
0.549 g K, we will run out of K first, so K is the limiting
reactant.
Note: It doesn't matter which reactant you start with;
the conclusion is the same. K is the limiting reactant.
>> Example 5a
How many grams of KCl can be made from 0.549 g potassium metal
mixed with 0.669 g of Cl2?
Solution with Method 1:
This is a continuation of Example 5. The same reaction and
quantities of reactants are used. From Example 5, we determined
that the limiting reactant was K and the balanced reaction is
2 K + Cl2 2
KCl
To solve the problem, we use the stoichiometry method, being
careful to start with the limiting reactant.
>>> Method 2
In method 2 you calculate the amount of product with both reactants
and choose the correct answer of the two, which is the lowest
answer.
>> Example 5b
How many grams of KCl can be made from 0.549 g potassium metal
mixed with 0.669 g of Cl2?
Solution with Method 2:
Step 1. Balance the equation.
2 K + Cl2 2
KCl
Steps 2-4. Assuming K is the limiting reactant.
Steps 2-4. Assuming Cl2 is the limiting reactant.
Choose the right answer, which is the lowest one—1.05
g KCl.
>> Example 6
How many grams of copper can you make from 0.094 mL of zinc
(density = 7.13 g/mL) and 7.74 x 1022
molec of copper(II) sulfate?
Zn + CuSO4 ZnSO4
+ Cu
Solution with Method 1:
Pick a reactant and compare to the other one.
The reaction is balanced. You can get to grams from milliliters
with density.
| 0.094 mL Zn |
|
|
|
|
|
|
|
| 6.022 x 1023 molec |
|
| 1 mol Cu |
|
= |
6.4 x 1021 molec CuSO4 |
That is less than the amount given, 7.74 x 1022
molec CuSO4, so zinc is the limiting reactant.
To solve the problem start with zinc and ignore the copper
sulfate.
Solution with Method 1:
The reaction is balanced.
Solve with both reactants.
| 7.74 x 1022 molec CuSO4 |
|
| 1 mol CuSO4 |
|
| 6.022 x 1023 molec |
|
|
|
|
|
= |
8.17 g Cu |
The lower answer is the correct answer, so the correct answer
is 0.65 g Cu.
>> Determining Actual,
Theoretical, and Percent Yields
Theoretical yield is the mass of product determined from a stoichiometric
calculation. In other words, it is the value determined by starting
with the limiting reactant and using the molar ratio from the balanced
chemical reaction.
For several reasons, the theoretical yield is also the maximum
yield of any reaction. The actual yield will be somewhat less. How
much less can only be determined experimentally, but it will be
proportional to the theoretical yield. This can be expressed as
percent yield.
| percent yield |
= |
| actual yield |
|
| theoretical yield |
|
|
100 |
The key to solving these problems is to identify how each value
fits into this relationship. Any units for mass can be used, but
the units must be the same for actual and theoretical yield.
>> Example 7
What is the percent yield of zinc chloride when 0.2583 g of ZnO
with excess HCl produces 0.2999 g ZnCl2.
ZnO + HCl H2O
+ ZnCl2
Solution:
| % yield |
= |
|
|
100 |
The question asks for percent yield; the product discussed is
ZnCl2. Its actual yield = 0.2999 g.
To find the theoretical yield, stoichiometry is used.
Step 1. Balance the reaction.
ZnO + 2 HCl H2O
+ ZnCl2
Steps 2. Start with the limiting reactant and convert to moles.
The limiting reactant is ZnO. The problem itself tells you that
HCl is in excess, therefore, ZnO is the limiting reactant by default.
Steps 2-4. Limiting reactant moles
LR moles product grams
product.
| 0.2583 g ZnO |
|
|
|
|
|
|
= |
0.4326 g ZnCl2 |
| |
|
|
|
|
|
|
= |
theoretical yield |
| % yield |
= |
|
|
100 |
= |
69.32% |
>> Example 8
The reaction of 0.213 g Mn(NO3)2 and 0.407
g K2CO3 produces MnCO3 with a
percent yield of 95.6%. How many grams of manganese(II) carbonate
will be made?
Mn(NO3)2 + K2CO3
MnCO3 + KNO3
Solution:
This time percent yield is given and actual yield is requested.
Therefore theoretical yield is needed. Since a limited quantity
of two reactants is given in the problem, the limiting reactant
must be determined. Using method 2 from the section on limiting
reactants, the amount of product is determined for each reactant.
Balance the equation.
Mn(NO3)2 + K2CO3
MnCO3 + 2 KNO3
Amount of product based on manganese(II) nitrate:
| 0.213 g |
|
|
|
| 1 mol MnCO3 |
|
| 1 mol Mn(NO3)2 |
|
|
|
= |
0.137 g |
Amount of product based on potassium carbonate:
Since manganese(II) nitrate makes a smaller amount of product,
it must be the limiting reactant and 0.137 g MnCO3
is the theoretical yield.
| % yield |
= |
|
|
100 |
| 95.6% |
= |
|
|
100 |
| 0.956(0.137) |
= |
actual yield |
| 0.131 g |
= |
actual yield |
>> Example 9
The following reaction produces iron metal at 74.2% yield. How
much Fe3O4 is needed to produce 2.35 kg
of Fe?
Fe3O4 + CO
Fe + CO2
Solution:
In this problem, the 2.35 kg of Fe is the actual yield. Percent
yield is also given, and the question asks about a reactant. Since
products and reactants are related theoretically by stoichiometry,
the theoretical yield is used to determine the amount of reactant.
| 74.2% |
= |
|
|
100 |
| 0.742 |
= |
|
|
|
| 0.742(theoretical yield) |
= |
2.35 kg |
| theoretical yield |
= |
3.17 kg |
Since we are looking for a reactant, the limiting reactant issue
is not relevant.
Balance the reaction.
Fe3O4 + 4 CO
3 Fe + 4 CO2
| 3.17 kg Fe |
|
|
|
|
|
|
|
|
= |
4380.97 g |
| |
|
|
|
|
|
|
|
|
= |
4.38 x 103 g |
| 4.38 x 103 g |
|
|
= |
4.38 kg |
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