Can you represent the oxidation of #"1-butylene"#?

The usual rigmarole is to balance the carbons as carbon dioxide, then balance the hydrogens as water, and then balance the oxygens.........you can get very good at this stoichiometry, and remember that you do this already every time you get change from a large bill....

For an alkanes, say butane, we follow the same rigmarole....

#C_4H_10(g) + 13/2O_2(g)rarr4CO_2(g) + 5H_2O(l)#

And thus here I took 4 equiv carbon dioxide, and then five equiv of water, and then balanced up the required dioxygens. You could double the entire equation to give....

#2C_4H_10(g) + 13O_2(g)rarr8CO_2(g) + 10H_2O(l)#

I think the arithmetic is easier in the former case.... Anyway try it out for #C_2H_6#, #H_2C=CHCH_3#, and #H_3C-CH_2OH#; these are probably all reasonable questions at your current level.

The formula for butene is

#C_4H_8#

You are going to bun it with plenty of oxygen:

#C_4H_8 + (n_1)O_2#

The products are carbon dioxide and water:

#C_4H_8 + (n_1)O_2 to (n_2)CO_2+ (n_3)H_2O#

Solved for the 3 variables:

#C_4 = (n_2)C#

#n_2 = 4#

#H_8 = (n_3)H_2#

#n_3=4#

#(n_1)O_2 = (4)O_2+(4)O#

#(n_1)O_2 = (4)O_2+(2)O_2#

#n_1 = 6#

The balanced equation is:

#C_4H_8 + 6O_2 to 4CO_2+ 4H_2O#

Unbalanced equation is:

#C_4H_6+O_2 -> CO_2+ H_2O#

First we can put #4# in front of #CO_2# to balance the number of carbon atoms:

#C_4H_6+O_2 -> 4CO_2+ H_2O#

Now we can put #3# next to water to balance hydrogene:

#C_4H_6+O_2 -> 4CO_2+ 3H_2O#

Hydrogen and carbon are now balanced, but not oxygen. On the right side there are #11# atoms, on the left side #2#. To balance it we have to put #5.5# on the left side:

#C_4H_6+5.5O_2 -> 4CO_2+ 3H_2O#

The equation is now balanced, but it is not correct to put fractions as the coefficients in a chemical reactions. To keep the equation ballanced and make all coefficients integer we have to multiply all coefficients by #2#

#2C_4H_6+11O_2 -> 8CO_2+ 6H_2O#