Balance the equation below.
C3H4 + O2 => H2O + CO2
Step 1) Start with C. We have 3carbon
atoms on the left and 1 on the right. So place a 3in front of the CO2
on the right.
C3H4 + O2
=> H2O + 3CO2
We now have 3 carbon atoms on both sides.
Step 2) Continue with H. We have 4 hydrogen atoms on the left and 2 on the
right. So place a 2 in front of the H2O on the rightt.
C3H4 + O2 => 2H2O + 3CO2
We now have 6 hydrogen atoms on both sides.
Step 3) Continue with O. We have
2 atoms of oxygen on the left and 8 on the right. So place a 4 in front of
the O2 on the leftt.
C3H4 + 4O2 => 3H2O + 3CO2
No fractions are required in this example.
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Balance the equation below.
HCl + O2 => Cl2 + H2O
Step 1) Start with Cl. We have
1 chlorine atom on the left and 2on the right. So place a 2 in front of the
HCl on the right.
2HCl + O2 => Cl2
+ H2O
We now have 2 chlorine atoms on both sides.
Step 2) Continue with H. We have 2 hydrogen atoms now on the left and 2 on
the right. The equation is so far balanced for hydrogen atoms.
2HCl + O2 => Cl2 + H2O
Step 3) Continue with O. We have
2 atoms of oxygen on the left and 1 on the right. So place a 1/2 in front
of the O2 on the leftt.
2HCl + 1/2O2 => Cl2 + H2O
Step 4) Multiply the entire equation
by the lowest number necessary to change all the fractions to a whole number.
(2HCl + 1/2O2 => Cl2 +
H2O ) X 2
=> 4HCl + O2 => 2Cl2
+ 2H2O --- This is a balanced equation.
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Balance the equation below.
H2S + O2 => SO2 + H2O
Step 1) Start with H. We have 2
hydrogen atoms on the left and 2 on the right. So far the equation is balanced
for hydrogen.
H2S + O2
=> SO2 + H2O
Step 2) Continue with S. We have 1 sulfur atom now on the left and 1 on the
right. So far the equation is balanced for hydrogen.
H2S + O2
=> SO2 + H2O
Step 3) Continue with O. We have
2 atoms of oxygen on the left and 3on the right. So place a 3/2 in front of
the O2 on the left.
H2S + 3/2O2 => SO2 + H2O
Step 4) Multiply the entire equation
by the lowest number necessary to change all the fractions to a whole number.
(H2S +3/2 O2 => SO2
+ H2O ) X 2
=> 2H2S + 3O2 => 2SO2
+ 2H2O --- This is a balanced equation.
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Balancing
chemical equations |
|
Balancing a chemical
equation can sometimes prove very difficult. The method shown below uses
fractions and may prove to be useful for some equations. 3Br2 + 8NH3 =>6NH4Br + N2 The number of each atom can be calculated by first multiplying the prefix in front of the compound and the subscript on the particular atom we are interested in finding. Example, to
calculate the number of hydrogen atoms on the left of the equation we
multiply the prefix (8) by the subscript (3) to get 8 X 3 = 24 hydrogen
atoms. |
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Balance the equation below. Br2 + NH3 =>NH4Br + N2 Step 1) Start with Br. We have
2 bromine atoms on the left and 1 on the right. So place a 1/2 in front
of the Br2 on the left.
Step 3) Continue with N. We
have 4/3 atoms on the left and 4 on the right. So place a 1/6 in front
of the N2 on the right. Step 4) Multiply the entire
equation by the lowest number necessary to change all the fractions to
a whole number. Using this method we sometimes refer to an equation as having a fraction of an atom. This is not the case and we multiply through by a number to remove fractions. Example 2 - Balance the equation below. C2H6+ O2 => CO2+ H2O Step 1) Start with C. We have
2 carbon atoms on the left and 1 on the right. So place a 2 in front of
the CO2 on the right.
Step 3) Continue with O. We
have 2 atoms of oxygen on the left and 7 on the right. So place a 7/2
in front of the O2 on the left. Step 4) Multiply the entire
equation by the lowest number necessary to change all the fractions to
a whole number. Try these |
|
| C3H4 + O2 → H2O + CO2 | Solution |
| HCl + O2 → Cl2 + H2O | Solution |
| H2S + O2 → SO2 + H2O | Solution |
| H2O + Al2S3 → H2S + Al(OH)3 | |
| Worksheet Solutions | |