Question

The general equation for the complete combustion of a hydrocarbon is
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SEP Analyze Data Complete each table to show the number of atoms as you balance each of the following equations for the combustion of hydrocarbons. Use the first reaction and table as an example. First count the atoms on each side of the reaction. As you balance the equation, cross out the original number and add the new number.



Atom Type
Reactants
Products
C atoms
1
1
H atoms
4

O atoms



general equation: (unbalanced)
balanced equation:
Atom Type
Reactants
Products
C atoms




H atoms




O atoms

Answer 1

1) General (unbalanced)
C_xH_y + O_2 → CO_2 + H_2O

Balanced (symbolic)
C_xH_y + (x + y/4) O_2 → x CO_2 + (y/2) H_2O
(If y is odd, multiply every coefficient by 2 to remove 1/2 coefficients.)

Example (methane CH4)
Balanced equation:
CH4 + 2 O2 → CO2 + 2 H2O

Example atom count table (filled)

Atom Type | Reactants | Products
C atoms | 1 | 1
H atoms | 4 | 4
O atoms | 4 (2×O2) | 4 (2 from CO2 + 2 from 2 H2O)

General atom count table (symbolic)

Atom Type | Reactants | Products
C atoms | x | x
H atoms | y | y
O atoms | 2·(x + y/4) = 2x + y/2 | 2x + y/2

(You can multiply the whole balanced equation by 2 when y is odd to get integer coefficients.)