Download Writing a Chemical Equation

Writing a Chemical Equation
Basic Theory
Having a balanced chemical equation is often vital to solving a chemistry question. The concept is to
express a reaction where product(s) come together to form one or more reactants. The basic rules are
simple.
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Each element must have the same number on both sides – Each atom entering a reaction must
exit the reaction.
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The total charge must be the same on both sides – Charges change (usually) by the transfer of
electrons. An atom that loses an electron (becoming more positive) must give it to an atom that
becomes more negative.
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Ionic compounds have an equal number of positive and negative ions – It is opposite charges
that glue ionic compounds together, so we need one of each. Also unless we are dealing with
lone ions the total charge, by definition, must be zero.
●
All numbers are to be the smallest integers possible (usually) – Having, say, ½ an atom
involved in a reaction is not physically realistic. And for simplicity we want the smallest
numbers possible.
Reaction types
Here are some types of reactions:
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➢
➢
Combustion – Reaction with oxygen
Single replacement – A lone atom kicks out an ion in a second compound and takes its place.
Double replacement – Two ions swap
Practice Set: Identify the reaction type (solutions at the end)
1. 2NaCl(aq) + Mg(OH)2(aq) → 2NaOH(aq) + MgCl2(s)
2. CH4(g) + 2O2(l) → CO2(g) + 2H2O(l)
3. Mg(s) + Fe2(SO4)3(aq) → Fe(s) + MgSO4(aq)
4. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
5. F2(g) + Li2S(aq) → 2LiF(aq) + S(s)
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Balancing Ions
When a compound is composed of ions the total charge is to equal zero. Below highlights a procedure
with two examples being worked in parallel.
Step
Aluminum Bromide
Al+3 Br-
Identify ions & charges
Manganese (IV) Sulfide
Mn+4
S-2
Combine, swap superscripts and make into
subscripts. Ignore signs.
Al1 Br3
Mn2 S4
Reduce subscripts by division as possible.
Ignore all ones.
AlBr3
MnS2
Practice Set: Balance the following ionic compounds (solutions at the end)
6. Calcium Chloride
7. Copper (II) Phosphate
8. Silver Nitrate
9. Manganese (VI) Phosphide
Balancing an equation
Once an equation type is identified and any ionic compounds balanced, then one can balance the
equation as a whole. In summary form the steps are:
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Identify the reaction type if needed to fill in the products in the reaction. Use this to identify
what compound(s) will be produced. Also identify any ions involved with their charges.
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Balance the ions within compounds as needed. If you skip this step you might not be able to
balance the equation at all!
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Choose an atom/ion that occurs exactly once on both sides of the equation. Following the
similar procedure you use in balancing ionic compounds swap the subscripts (making them
coefficients out front of the compounds) and reduce.
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Given the assumed coefficients for your compounds, balance the remaining atoms/ions one at a
time. Repeat this step until all coefficients are known.
These steps will be illustrated on the next page as the reaction of Barium with Phosphoric acid (H3PO4)
is balanced.
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Step
Result
Identify reaction type
Single replacement, so Ba and H swap
Basic form of the equation with charges
Ba + H+ PO4-3 → H2 + Ba+2 PO4-3
Balance the ionic compounds
Ba + H3PO4 → H2 + Ba3(PO4)2
Balance your first compound by swapping
subscripts and placing as coefficients out front of
the compounds
Ba + 2H3PO4 → 3H2 + Ba3(PO4)2
Insure that each ion or atom is balanced one at a
time
Ba + 2H3PO4 → 3H2 + 1Ba3(PO4)2
(Balance PO4 next, but it's already good)
Repeat as needed
3Ba + 2H3PO4 → 3H2 + 1 Ba3(PO4)2
(Lastly the Ba)
(Here H is balanced first)
As a second illustration consider the combustion of Octane (the chief component of gasoline) C8H18.
This is useful because sometimes one encounters a case where one cannot balance a particular
compound using integers only. In this case one can simply multiply all previously determined
coefficients by an appropriate number.
Step
Result
Identify reaction type
Combustion
Basic form of the equation with charges
C8H18 + O2 → CO2 + H2O
Balance the ionic compounds
N/A
Balance your first compound by swapping
subscripts and placing as coefficients out front of
the compounds
1C8H18 + O2 → 8CO2 + H2O
Insure that each ion or atom is balanced one at a
time
1C8H18 + O2 → 8CO2 + 9H2O
(Balance the H next)
(Here C is balanced first)
Here the 25 O atoms on the right cannot fit into the 2C8H18 + O2 → 16CO2 + 18H2O
O2 on the left. Adjust the known coefficients
(Multiply known coefficients by 2)
Continue balancing
2C8H18 + 25O2 → 16CO2 + 18H2O
(Balance the O)
Note: We have used the coefficient one to balance our equations. However we typically do not include
one as a coefficient.
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Practice Set: Balance the following (solutions at the end)
10. Methane (CH4) reacting with oxygen gas
11. Calcium metal reacting with a solution of Silver Sulfate
12. Benzene (C6H6) and oxygen gas
13. Potassium metal reacting with water
14. Ammonium Chloride and Magnesium acetate reacting in solution
15. Rubidium Sulfide reacting with Iron (III) Carbonate in solution
16. Ethanol alcohol (C2H5OH) burning in oxygen
Solutions
1. 2NaCl(aq) + Mg(OH)2(aq) → 2NaOH(aq) + MgCl2(s)
2. CH4(g) + 2O2(l) → CO2(g) + 2H2O(l)
3. Mg(s) + Fe2(SO4)3(aq) → Fe(s) + MgSO4(aq)
4. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
5. F2(g) + Li2S(aq) → 2LiF(aq) + S(s)
6.
7.
8.
9.
Double Replacement
Combustion
Single Replacement
Combustion
Single Replacement
CaCl2
Cu3(PO4)2
AgNO3
MnP2
10. CH4 + 2O2 → CO2 + 2H2O
11. Ca + Ag2SO4 → 2Ag + CaSO4
12. 2C6H6 + 15O2 → 12CO2 + 6H2O
13. 2K + H2O → K2O + H2
14. 2NH4Cl + Mg(CH3CO2)2 → 2NH4CH3CO2 + MgCl2
15. 3Rb2S + Fe2(CO3)3 → 3Rb2CO3 + Fe2S3
16. C2H5OH + 3O2 → 2CO2 + 3H2O
Prepared by Kevin Gibson
www.mc.maricopa.edu/~kevinlg/supplements/
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