Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Fall 2008
	© R. Paselk 2008
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Chemical Equations

Equation Balancing by Inspection

The key here is to remember Conservation of Mass, that is there must be the same amount of stuff before and after the reaction. Since atoms are not created nor destroyed in chemical processes (mass is conserved), there must be the same nuimber of each kind of atoms on both sides of each equations. Let's give it a try.

Balance the equations below by providing the proper coefficients:

1. 1. Al_(s) + Fe₂O_{3 (s)} Fe_(s) + Al ₂O_{3 (s)}

   Looking at the reaction as written we see that there are different numbers of atoms on the two sides - mass is not conserved. We can make a table to keep track of the numbers of atoms on both sides:

   	Reactants	Products
   Al	1	2
   Fe	2	1
   O	3	3

   Looking at the table, oxygen is already balanced - both sides have the same number, and we simply need to multiply Al and Fe by 2 by adding the appropriate coeficients:

   2 Al_(s) + Fe₂O_{3 (s)} 2 Fe_(s) + Al ₂O_{3 (s)}

   The table of atoms now shows all are balanced:

   	Reactants	Products
   Al	2	2
   Fe	2	2
   O	3	3

    

   2. KClO_{3 (s)} + heat KCl_(s) + O_{2 (g)}

   Making a table to keep track of the numbers of atoms on both sides:

   	Reactants	Products
   K	1	1
   Cl	1	1
   O	3	2

   Looking at the table, oxygen is the only atom not balanced, but this time we must multiply KClO₃by 2 and O₂ by 3:

   2 KClO_{3 (s)} + heat KCl_(s) + 3 O_{2 (g)}

   Updating the table we get:

   	Reactants	Products
   K	2	1
   Cl	2	1
   O	6	6

   Addding a coefficient of 2 for KCl will then balance the equation:

   2 KClO_{3 (s)} + heat 2 KCl_(s) + 3 O_{2 (g)}

    

   3. C₄H_{10 (g)} + O_{2 (g)} CO_{2 (g)} + H₂O_(g)

   For combustion reactions it is usually easier to balance the oxygen last. Following this strategy, we can balance C by multiplying CO₂by 4:

C₄H_{10 (g)} + O_{2 (g)} 4 CO_{2 (g)} + H₂O_(g)

Carbon is now balanced, next balance H by multipying by 5 to give 10 hydrogens on each side:

C₄H_{10 (g)} + O_{2 (g)} 4 CO_{2 (g)} + 5 H₂O_(g)

Finally we can balance oxygen by adding up the number of oxygens on the right, 4(2) = 8 + 5(1) = 10 = 13. But each O₂ has 2 oxygens, so only need 13/2:

C₄H_{10 (g)} + 13/2 O_{2 (g)} 4 CO_{2 (g)} + 5 H₂O_(g)

This equation is now balanced - mass is conserved. However it is conventional to not use fractions. Multiplying by 2 gives integral coefficients and our final answer:

2 C₄H_{10 (g)} + 13 O_{2 (g)} 8 CO_{2 (g)} + 10 H₂O_(g)

4. N_{2 (g)} + H_{2 (g)} NH_{3 (g)}

Looking at the reaction as written we see that there are different numbers of nitrogen on the two sides, 2 on the reactant (right) side (as N₂) but only one on the product (left) side (as NH₃). Multiplying the ammonia (NH₃) by 2 then gives:

N_{2 (g)} + H_{2 (g)} 2 NH_{3 (g)}

Now looking at the hydrogens we see 2 hydrogen atoms on the reactant side and 2(3) = 6 hydrogen atoms on the product side. Multiplying the hydrogen (H₂) by three will give a balanced equation:

N_{2 (g)} + 3 H_{2 (g)} 2 NH_{3 (g)}

5. P_{4 (s)}+ F_{2 (g)} PF_{5 (g)}

Looking at the reaction as written we see that there are different numbers of phosphorus on the two sides, 4 on the reactant (right) side (as P₄)* but only one on the product (left) side. Multiplying the phosphorus pentafluoride (PF₅) by 4 then gives:

P_{4 (s)}+ F_{2 (g)} 4 PF_{5 (g)}

Now looking at the fluorines we see 2 fluorine atoms on the reactant side and 4(5) = 10 hydrogen atoms on the product side. Multiplying the fluorine (F₂) by 4 will give a balanced equation:

P_{4 (s)}+ 10 F_{2 (g)} 4 PF_{5 (g)}

*Phosphorus exists as a tetraatomic molecule in its elemental state.

© R A Paselk

Last modified 23 October 2008