Difference between revisions of "Chemical equation" - New World Encyclopedia

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A '''chemical equation''' is a symbolic representation of a [[chemical reaction]].<ref>[[IUPAC]] Compendium of Chemical Terminology </ref> The coefficients next to the symbols and formulae of entities are the absolute values of the [[stoichiometric coeff jk,,icient|stoichiometric numbers]].  The first chemical equation was diagrammed by [[Jean Beguin]] in 1615.  
A '''chemical equation''' is a mathematical method used to display the ‘before’, the ‘transition’, and ‘after’ stages for [[chemical reactions]], wherein substances are changed into other substances. The species to the left of the arrow represent the [[reactants]], the arrow represents the [[transition state|transition stage]], and the species to the right of the arrow represent the [[products]]. The four basic chemical equations are:
 
  
::'''''A &rarr; B'''''
 
  
::'''''A &rarr; B + C'''''
 
  
:'''''A + B &rarr; C'''''
+
== Balancing chemical equations ==
 +
In a [[chemical reaction]], the quantity of each element does not change.  Thus, each side of the equation must represent the same quantity of any particular element.  Also in case of net ionic reactions the same charge must be present on both sides of the hiddly unbalanced equation, one may balance it by changing the scalar number for each [[molecular formula]].
  
:'''''A + B &rarr; B + C'''''
+
Simple chemical equations can be balanced by inspection, that is, by trial and error.  Generally, it is best to balance the most complicated molecule first. Hydrogen and oxygen are usually balanced last.
  
For example, the combustion of [[methane]] in [[oxygen]] is:
+
Ex #1.  Na + O<sub>2</sub> → Na<sub>2</sub>O
  
:CH<sub>4</sub> + 2&nbsp;O<sub>2</sub> &rarr; CO<sub>2</sub> + 2&nbsp;H<sub>2</sub>O,
+
In order for this equation to be balanced, there must be an equal amount of Na on the left hand side as on the right hand side. As it stands now, there is 1 Na on the left but 2 Na's on the right. This problem is solved by putting a 2 in front of the Na on the left hand side:
  
and the [[reversible reaction]] of the [[Haber process]] is
+
:2Na + O<sub>2</sub> → Na<sub>2</sub>O
  
:N<sub>2(g)</sub> + 3H<sub>2(g)</sub> &harr; 2NH<sub>3(g)</sub> + &Delta;H.
+
In this there are 2 Na atoms on the left and 2 Na atoms on the right. In the next step the oxygen atoms are balanced as well. On the left hand side there are 2 O atoms and the right hand side only has one. This is still an unbalanced equation. To fix this  a 2 is added in front of the Na<sub>2</sub>O on the right hand side. Now the equation reads:
  
== Reading Chemical Equations ==
+
:2Na + O<sub>2</sub> → 2Na<sub>2</sub>O
  
* Each side of an equation represents a mixture of chemicals. The mixture is written as a set of [[molecular formula]]s, separated by '''+''' symbols.
+
Notice that the 2 on the right hand side is "distributed" to both the Na<sub>2</sub> and the O. Currently the left hand side of the equation has 2 Na atoms and 2 O atoms. The right hand side has 4 Na's total and 2 O's. Again, this is a problem, there must be an equal amount of each chemical on both sides. To fix this 2 more Na's are added on the left side. The equation will now look like this:
* Each formula is preceeded by an optional scaler number (if no scaler number is written, it is implied that the number is 1). The scaler numbers indicate the relative quantity of molecules in the reaction. For instance, the string '''2H<sub>2</sub>O + 3CH<sub>4</sub>''' represents a mixture containing 2 molecules of '''H<sub>2</sub>O''' for every 3 molecules of '''CH<sub>4</sub>'''.
+
 
* The two sides of the equation are separated by an arrow.  If the reaction is non-reversable, a right-arrow (→) is used, indicating that the left side represents the mixture of chemicals before the reaction, and the right side indicates the mixture after the reactionFor a [[reversible reaction]], a two-way arrow is used.
+
:4Na + O<sub>2</sub> → 2Na<sub>2</sub>O
 +
 
 +
This equation is a balanced equation because there is an equal number of atoms of each element on the left and right hand sides of the equation.
 +
 
 +
Ex #2.
 +
This equation is not balanced because there is an unequal amount of O's on both sides of the equation. The left hand side has 4 P's and the right hand side has 4 P's. So the P atoms are balanced. The left hand side has 2 O's and the right hand side has 10 O's.
 +
 
 +
:P<sub>4</sub> + O<sub>2</sub> → 2P<sub>2</sub>O<sub>5</sub>
 +
 
 +
To fix this unbalanced equation a 5 in front of the O<sub>2</sub> on the left hand side is added to make 10 O's on both sides resulting in
 +
 
 +
:P<sub>4</sub> + 5O<sub>2</sub> → 2P<sub>2</sub>O<sub>5</sub>
 +
 
 +
The equation is now balanced because there is an equal amount of substances on the left and the right hand side of the equation.
 +
 
 +
Ex #3.   C<sub>2</sub>H<sub>5</sub>OH + O<sub>2</sub> → CO<sub>2</sub> + H<sub>2</sub>O
  
'''Example: 4Na + O<sub>2</sub> &rarr; 2Na<sub>2</sub>O'''
+
This equation is more complex than the previous examples and requires more steps. The most complicated molecule here is C<sub>2</sub>H<sub>5</sub>OH, so balancing begins by placing the coefficient 2 before the CO<sub>2</sub> to balance the carbon atoms.
  
This equation represents a non-reversable reaction.  In this reaction, [[sodium]]('''Na''') and [[oxygen]]('''O<sub>2</sub>''') are converted to a single molecule, '''Na<sub>2</sub>O''' (containing 2 sodium atoms and 1 oxygen atom). We can also see that for every 4 sodium atoms at the beginning of the reaction, a single '''O<sub>2</sub>''' molecule will participate, and 2 '''Na<sub>2</sub>O''' molecules will result.
+
:C<sub>2</sub>H<sub>5</sub>OH + O<sub>2</sub> → 2CO<sub>2</sub> + H<sub>2</sub>O
  
== Balancing Chemical Equations ==
+
Since C<sub>2</sub>H<sub>5</sub>OH contains 6 hydrogen atoms, the hydrogen atoms can be balanced by placing 3 before the H<sub>2</sub>O:
  
In a [[chemical reaction]], the quantity of each element does not change. Thus, each side of the equation must represent the same quantity of any particular element.  When this is true, the equation is balanced.  Given an unbalanced equation, one may balance it by changing the scaler number for each [[molecular formula]].
+
:C<sub>2</sub>H<sub>5</sub>OH + O<sub>2</sub> → 2CO<sub>2</sub> + 3H<sub>2</sub>O
  
Simple chemical equations can be balanced by inspection, that is, by trial and error.  Generally, it is best to balance the most complicated molecule first.
+
Finally the oxygen atoms must be balanced. Since there are 7 oxygen atoms on the right and only 3 on the left, a 3 is placed before O<sub>2</sub>, to produce the balanced equation:
  
Let's look at a few examples and walk through it:
+
:C<sub>2</sub>H<sub>5</sub>OH + 3O<sub>2</sub> →  2CO<sub>2</sub> + 3H<sub>2</sub>O
  
'''Ex #1.  Na + O<sub>2</sub> &rarr; Na<sub>2</sub>O'''
+
=== Linear system balancing ===
  
In order for this equation to be balanced, there must be equal amount of Na on the left hand side as on the right hand side. As it stands now, there is 1 Na on the left but 2 Na's on the right. We solve this problem by putting a 2 in front of the Na on the left hand side, Like this:
+
In reactions involving many compounds, equations may then be balanced using an algebraic method, based on solving set of linear equations.
  
    2Na + O<sub>2</sub> &rarr; Na<sub>2</sub>O
+
1. Assign variables to each coefficient. (Coefficients represent both the basic unit and mole ratios in balanced equations.):
  
So now we see, there are 2 Na's on the left and 2 Na's on the right. But what about the O's? We now must check to see if the O's are balanced on both sides of the equation. On the left hand side there are 2 O's and the Right hand side only has one. This is still an unbalanced equation. To fix this we must put a 2 in front of the Na<sub>2</sub>O on the right hand side. Now our equation reads:
+
*'''a K<sub>4</sub>Fe(CN)<sub>6</sub> + b H<sub>2</sub>SO<sub>4</sub> + c H<sub>2</sub>O → d K<sub>2</sub>SO<sub>4</sub> + e FeSO<sub>4</sub> + f (NH<sub>4</sub>)<sub>2</sub>SO<sub>4</sub> + g CO'''
  
    2Na + O<sub>2</sub> &rarr; 2Na<sub>2</sub>O
+
2. There must be the same quantities of each atom on each side of the equation. So, for each element, count its atoms and let both sides be equal.
  
Notice that the 2 on the right hand side is "distributed" to both the Na<sub>2</sub> and the O. Currently the left hand side of the equation has 2 Na's and 2O's. The right hand side has 4 Na's total and 2 O's. Again, this is a problem, there must be an equal amount of each chemical on both sides. To fix this let's add 2 more Na's on the left side. The equation will now look like this:
+
*'''K:  4a = 2d'''
 +
*'''Fe: 1a = 1e'''
 +
*'''C:  6a = g'''
 +
*'''N:  3a = f'''
 +
*'''H:  2b+2c = 8f'''
 +
*'''S:  b = d+e+f'''
 +
*'''O: 4b+c = 4d+4e+4f+g'''
  
    4Na + O<sub>2</sub> &rarr; 2Na<sub>2</sub>O
+
3. Solve the system (Direct substitution is usually the best way.)
  
So now, as you can see, we have a balanced equation because there is an equal amount of element's on the left and right hand sides of the equation.
+
*'''d=2a'''
 +
*'''e=a'''
 +
*'''g=6a'''
 +
*'''f=3a'''
 +
*'''b=6a'''
 +
*'''c=6a'''
  
 +
which means that all coefficients depend on a parameter a, just choose a=1 (a number that will make all of them small whole numbers), which gives:
  
'''Ex #2.    P<sub>4</sub> + O<sub>2</sub> &rarr; P<sub>4</sub>O<sub>10</sub>'''
+
*'''a=1 b=6 c=6 d=2 e=1 f=3 g=6'''
  
This equation is not balanced because there is an unequal amount of O's on both sides of the equation. The left hand side has 4 P's and the right hand side has 4 P's. So the P's balance. Let's look at the O's. The left hand side has 2 O's and the right hand side has 10 O's. To fix this unbalanced equation we must put a 5 in front of the O<sub>2</sub> on the left hand side to make 10 O's on both sides. Let's take a look at what this looks like:
+
4. And the balanced equation at last:{{IPA|}}
  
        P<sub>4</sub> + 5O<sub>2</sub> &rarr; P<sub>4</sub>O<sub>10</sub>
+
*'''K<sub>4</sub>Fe(CN)<sub>6</sub> + 6 H<sub>2</sub>SO<sub>4</sub> + 6 H<sub>2</sub>O → 2 K<sub>2</sub>SO<sub>4</sub> + FeSO<sub>4</sub> + 3 (NH<sub>4</sub>)<sub>2</sub>SO<sub>4</sub> + 6 CO'''
  
The equation is now balanced because there is an equal amount of substances on the left and the right hand side of the equation.
+
To speed up the process, one can combine both methods to get a more practical algorithm:
  
 +
1. Identify elements which occur in one compound in each member. (This is very usual.)
  
'''Ex #3.   C<sub>2</sub>H<sub>5</sub>OH + O<sub>2</sub> &rarr;  CO<sub>2</sub> + H<sub>2</sub>O'''
+
2. Start with the one among those which has a big index (this will help to keep working with integers), and assign a variable, such as a.
  
This equation is more complex than the previous examples; it will take a few steps. The most complicated molecule here is C<sub>2</sub>H<sub>5</sub>OH, so we begin by placing the coefficient 2 before the CO<sub>2</sub> to balance the carbon atoms.
+
*'''a K<sub>4</sub>Fe(CN)<sub>6</sub> + H<sub>2</sub>SO<sub>4</sub> + H<sub>2</sub>O → K<sub>2</sub>SO<sub>4</sub> + FeSO<sub>4</sub> + (NH<sub>4</sub>)<sub>2</sub>SO<sub>4</sub> + CO'''
  
        C<sub>2</sub>H<sub>5</sub>OH + O<sub>2</sub> &rarr;  2CO<sub>2</sub> + H<sub>2</sub>O
+
3. K<sub>2</sub>SO<sub>4</sub> has to be 2a (because of K), and also, FeSO<sub>4</sub> has to be 1a (because of Fe), CO has to be 6a (because of C) and (NH<sub>4</sub>)<sub>2</sub>SO<sub>4</sub> has to be 3a (because of N). This removes the first four equations of the system. It is already known that whatever the coefficients are, those proportions must hold:
  
Since C<sub>2</sub>H<sub>5</sub>OH contains 6 hydrogen atoms, the hydrogen atoms can be balanced by placing 3 before the H<sub>2</sub>O, like this:
+
*'''a K<sub>4</sub>Fe(CN)<sub>6</sub> + H<sub>2</sub>SO<sub>4</sub> + H<sub>2</sub>O → 2a K<sub>2</sub>SO<sub>4</sub> + a FeSO<sub>4</sub> + 3a (NH<sub>4</sub>)<sub>2</sub>SO<sub>4</sub> + 6a CO'''
  
        C<sub>2</sub>H<sub>5</sub>OH + O<sub>2</sub> &rarr;  2CO<sub>2</sub> + 3H<sub>2</sub>O
+
4. One can continue by writing the equations now (and having simpler problem to solve) or, in this particular case (although not so particular) one could continue by noticing that adding the Sulfurs yields 6a for H<sub>2</sub>SO<sub>4</sub> and finally by adding the hydrogens (or the oxygens) one can find the lasting 6a for H<sub>2</sub>SO<sub>4</sub>.
  
Finally we balance the oxygen atoms. Since there are 7 oxygen atoms on the right and only 3 on the left, we balance it by placing a 3 before O<sub>2</sub>, to produce the balanced equation:
+
5. Again, having a convenient value for a (in this case 1 will do, but if a results in fractional values in the other coefficients, one would like to cancel the denominators) The result is
  
        C<sub>2</sub>H<sub>5</sub>OH + 3O<sub>2</sub> &rarr;  2CO<sub>2</sub> + 3H<sub>2</sub>O
+
*'''K<sub>4</sub>Fe(CN)<sub>6</sub> + 6 H<sub>2</sub>SO<sub>4</sub> + 6 H<sub>2</sub>O → 2 K<sub>2</sub>SO<sub>4</sub> + FeSO<sub>4</sub> + 3 (NH<sub>4</sub>)<sub>2</sub>SO<sub>4</sub> + 6 CO'''
  
== See also ==  
+
== Reading chemical equations ==
 +
When reading a chemical equation there are some points to consider.
 +
* Each side of an equation represents a mixture of chemicals.  The mixture is written as a set of [[molecular formula]]s, separated by '''+''' symbols.
 +
* Each formula is preceded by an optional '''scalar number''' (if no scalar number is written, it is implied that the number is 1).  The scalar numbers indicate the relative quantity of molecules in the reaction.  For instance, the string 2H<sub>2</sub>O + 3CH<sub>4</sub> represents a mixture containing 2 molecules of H<sub>2</sub>O for every 3 molecules of CH<sub>4</sub>.
 +
* The two sides of the equation are separated by an arrow.  If the reaction is non-reversible, a right-arrow (→) is used, indicating that the left side represents the mixture of chemicals before the reaction, and the right side indicates the mixture after the reaction.  For a [[reversible reaction]], a two-way arrow is used. For example the equation 4Na + O<sub>2</sub> → 2Na<sub>2</sub>O represents a non-reversible reaction.  In this reaction, [[sodium]] (Na) and [[oxygen]] (O<sub>2</sub>) are converted to a single molecule, Na<sub>2</sub>O (containing 2 sodium atoms and 1 oxygen atom).  We can also see that for every 4 sodium atoms at the beginning of the reaction, a single O<sub>2</sub> molecule will participate, and 2 Na<sub>2</sub>O molecules will result.
 +
* A chemical equation does not imply that all reactants are consumed in a [[chemical process]]. For instance a [[limiting reactant]] determines how far a reaction can go.
 +
* In an [[ionic equation]] balancing of charge also takes place. In a [[full equation]] all reactants are written as molecules.
  
*[[Full equation]]
+
==References==
*[[Ionic equation]]
+
<references />
*[[Limiting reagent]]
 
  
 
== External links ==
 
== External links ==
 
+
* [http://funbasedlearning.com/chemistry/chemBalancer/default.htm Classic Chembalancer] - Play Chembalancer, a free online game at FunBasedLearning.com, to learn how to balance equations by inspection
 
* [http://sciencesoft.at/index.jsp?link=solve&lang=en Online calculator], determines of the coefficients of a chemical equation
 
* [http://sciencesoft.at/index.jsp?link=solve&lang=en Online calculator], determines of the coefficients of a chemical equation
 
+
* [http://adomas.org/bceq/ Free chemical equation balancer], solves chemical equation, works online, low bandwidth
[[Category:Physical sciences]]
+
* [http://www.chembuddy.com/?left=EBAS&right=balancing-chemical-equations Chemical equation balancing program] - works off-line, calculates stoichiometry and limiting reagents, balances charge.
[[Category:Chemistry]]
+
* [http://www.webqc.org/balance.php Online Chemical Equation Balancer] Balances equation of any chemical reaction (full or half-cell) in one click.
 +
* [http://mgccl.com/2007/09/20/balance-chemical-equations Balance chemical equations] Teaches how to balance chemical equations
 +
[[Category:Stoichiometry]]
 
[[Category:Equations]]
 
[[Category:Equations]]
  
 
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[[el:Χημική εξίσωση]]
 
[[es:Ecuación química]]
 
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[[fr:Équation chimique]]
 
[[fr:Équation chimique]]
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[[hi:रासायनिक समीकरण]]
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Revision as of 23:28, 29 August 2008

A chemical equation is a symbolic representation of a chemical reaction.[1] The coefficients next to the symbols and formulae of entities are the absolute values of the stoichiometric numbers. The first chemical equation was diagrammed by Jean Beguin in 1615.


Balancing chemical equations

In a chemical reaction, the quantity of each element does not change. Thus, each side of the equation must represent the same quantity of any particular element. Also in case of net ionic reactions the same charge must be present on both sides of the hiddly unbalanced equation, one may balance it by changing the scalar number for each molecular formula.

Simple chemical equations can be balanced by inspection, that is, by trial and error. Generally, it is best to balance the most complicated molecule first. Hydrogen and oxygen are usually balanced last.

Ex #1. Na + O2 → Na2O

In order for this equation to be balanced, there must be an equal amount of Na on the left hand side as on the right hand side. As it stands now, there is 1 Na on the left but 2 Na's on the right. This problem is solved by putting a 2 in front of the Na on the left hand side:

2Na + O2 → Na2O

In this there are 2 Na atoms on the left and 2 Na atoms on the right. In the next step the oxygen atoms are balanced as well. On the left hand side there are 2 O atoms and the right hand side only has one. This is still an unbalanced equation. To fix this a 2 is added in front of the Na2O on the right hand side. Now the equation reads:

2Na + O2 → 2Na2O

Notice that the 2 on the right hand side is "distributed" to both the Na2 and the O. Currently the left hand side of the equation has 2 Na atoms and 2 O atoms. The right hand side has 4 Na's total and 2 O's. Again, this is a problem, there must be an equal amount of each chemical on both sides. To fix this 2 more Na's are added on the left side. The equation will now look like this:

4Na + O2 → 2Na2O

This equation is a balanced equation because there is an equal number of atoms of each element on the left and right hand sides of the equation.

Ex #2. This equation is not balanced because there is an unequal amount of O's on both sides of the equation. The left hand side has 4 P's and the right hand side has 4 P's. So the P atoms are balanced. The left hand side has 2 O's and the right hand side has 10 O's.

P4 + O2 → 2P2O5

To fix this unbalanced equation a 5 in front of the O2 on the left hand side is added to make 10 O's on both sides resulting in

P4 + 5O2 → 2P2O5

The equation is now balanced because there is an equal amount of substances on the left and the right hand side of the equation.

Ex #3. C2H5OH + O2 → CO2 + H2O

This equation is more complex than the previous examples and requires more steps. The most complicated molecule here is C2H5OH, so balancing begins by placing the coefficient 2 before the CO2 to balance the carbon atoms.

C2H5OH + O2 → 2CO2 + H2O

Since C2H5OH contains 6 hydrogen atoms, the hydrogen atoms can be balanced by placing 3 before the H2O:

C2H5OH + O2 → 2CO2 + 3H2O

Finally the oxygen atoms must be balanced. Since there are 7 oxygen atoms on the right and only 3 on the left, a 3 is placed before O2, to produce the balanced equation:

C2H5OH + 3O2 → 2CO2 + 3H2O

Linear system balancing

In reactions involving many compounds, equations may then be balanced using an algebraic method, based on solving set of linear equations.

1. Assign variables to each coefficient. (Coefficients represent both the basic unit and mole ratios in balanced equations.):

  • a K4Fe(CN)6 + b H2SO4 + c H2O → d K2SO4 + e FeSO4 + f (NH4)2SO4 + g CO

2. There must be the same quantities of each atom on each side of the equation. So, for each element, count its atoms and let both sides be equal.

  • K: 4a = 2d
  • Fe: 1a = 1e
  • C: 6a = g
  • N: 3a = f
  • H: 2b+2c = 8f
  • S: b = d+e+f
  • O: 4b+c = 4d+4e+4f+g

3. Solve the system (Direct substitution is usually the best way.)

  • d=2a
  • e=a
  • g=6a
  • f=3a
  • b=6a
  • c=6a

which means that all coefficients depend on a parameter a, just choose a=1 (a number that will make all of them small whole numbers), which gives:

  • a=1 b=6 c=6 d=2 e=1 f=3 g=6

4. And the balanced equation at last:

  • K4Fe(CN)6 + 6 H2SO4 + 6 H2O → 2 K2SO4 + FeSO4 + 3 (NH4)2SO4 + 6 CO

To speed up the process, one can combine both methods to get a more practical algorithm:

1. Identify elements which occur in one compound in each member. (This is very usual.)

2. Start with the one among those which has a big index (this will help to keep working with integers), and assign a variable, such as a.

  • a K4Fe(CN)6 + H2SO4 + H2O → K2SO4 + FeSO4 + (NH4)2SO4 + CO

3. K2SO4 has to be 2a (because of K), and also, FeSO4 has to be 1a (because of Fe), CO has to be 6a (because of C) and (NH4)2SO4 has to be 3a (because of N). This removes the first four equations of the system. It is already known that whatever the coefficients are, those proportions must hold:

  • a K4Fe(CN)6 + H2SO4 + H2O → 2a K2SO4 + a FeSO4 + 3a (NH4)2SO4 + 6a CO

4. One can continue by writing the equations now (and having simpler problem to solve) or, in this particular case (although not so particular) one could continue by noticing that adding the Sulfurs yields 6a for H2SO4 and finally by adding the hydrogens (or the oxygens) one can find the lasting 6a for H2SO4.

5. Again, having a convenient value for a (in this case 1 will do, but if a results in fractional values in the other coefficients, one would like to cancel the denominators) The result is

  • K4Fe(CN)6 + 6 H2SO4 + 6 H2O → 2 K2SO4 + FeSO4 + 3 (NH4)2SO4 + 6 CO

Reading chemical equations

When reading a chemical equation there are some points to consider.

  • Each side of an equation represents a mixture of chemicals. The mixture is written as a set of molecular formulas, separated by + symbols.
  • Each formula is preceded by an optional scalar number (if no scalar number is written, it is implied that the number is 1). The scalar numbers indicate the relative quantity of molecules in the reaction. For instance, the string 2H2O + 3CH4 represents a mixture containing 2 molecules of H2O for every 3 molecules of CH4.
  • The two sides of the equation are separated by an arrow. If the reaction is non-reversible, a right-arrow (→) is used, indicating that the left side represents the mixture of chemicals before the reaction, and the right side indicates the mixture after the reaction. For a reversible reaction, a two-way arrow is used. For example the equation 4Na + O2 → 2Na2O represents a non-reversible reaction. In this reaction, sodium (Na) and oxygen (O2) are converted to a single molecule, Na2O (containing 2 sodium atoms and 1 oxygen atom). We can also see that for every 4 sodium atoms at the beginning of the reaction, a single O2 molecule will participate, and 2 Na2O molecules will result.
  • A chemical equation does not imply that all reactants are consumed in a chemical process. For instance a limiting reactant determines how far a reaction can go.
  • In an ionic equation balancing of charge also takes place. In a full equation all reactants are written as molecules.

References
ISBN links support NWE through referral fees

  1. IUPAC Compendium of Chemical Terminology

External links

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