Write out the balanced reactions for combustion of methanol, ethanol, propanol, butanol, and pentanol?

I will do pentanol, and the common sequence is (i) to balance the carbons; (ii) balance the hydrogens; and (iii) balance the oxygens....

#C_5H_11OH(s)+15/2O_2(g)rarr5CO_2(g) + 6H_2O(l) + Delta#

Odd-numbered alcohols require non-integral stoichiometric coefficients.....you can always remove the #1/2# coefficient by doubling the equation.

#color(green)(2)C_n H_(2n+1)OH(l) + color(green)(3n)O_2(g) -> color(green)(2n)CO_2(g) + color(green)((2n+2))H_2O(l)#

#n = 1, 2, 3 . . . # is the number of carbon atoms in the alcohol.

If #n# is odd, you are done. If #n# is even, divide by #2#.

Each alcohol follows the same general formula, #C_n H_(2n+1)OH#, or #C_n H_(2n+2)O#, where #n = 1, 2, 3, . . . #.

If we balance one in general, presumably you can use it to balance the particular reactions above...

#C_n H_(2n+2)O(l) + O_2(g) -> CO_2(g) + H_2O(l)#

#1)# Balance the carbons

#C_n H_(2n+2)O(l) + O_2(g) -> nCO_2(g) + H_2O(l)#

We now have #n# carbon atoms on each side.

#2)# Balance the hydrogens

Given #2# hydrogen atoms on the righthand side, we solve

#2n+2 = 2m#,

where #m ne n#. Hence, #m = n+1#, and we have:

#C_n H_(2n+2)O(l) + O_2(g) -> nCO_2(g) + (n+1)H_2O(l)#

We now have #2n+2# hydrogen atoms on each side.

#3)# Balance the oxygens

Balancing the oxygen atoms should be simple... except our alcohol has an oxygen too...

The #n+1# makes for a difficult balancing act, so let's double everything to see a little better, and multiply #O_2# by #x# to balance using that next, #x ne n#.

#2C_n H_(2n+2)O(l) + 2xO_2(g) -> 2nCO_2(g) + (2n+2)H_2O(l)#

Now we have #6n+2# oxygen atoms on the right-hand side, and #2 + 4x# on the left-hand side. Thus, we construct another quick balancing equation:

#2 + 2(2x) = 2(2n) + (2n + 2)#

#=> x = 3/2n#

Plug in #x# to get:

#color(blue)(2C_n H_(2n+2)O(l) + 3nO_2(g) -> 2nCO_2(g) + (2n+2)H_2O(l))#

Now as a check, we have:

• #2 + 6n# oxygen atoms on the left, and #4n + 2n + 2# oxygen atoms on the right. #color(blue)(sqrt"")#
• #2n# carbon atoms on the left and right sides. #color(blue)(sqrt"")#
• #2(2n+2)# hydrogen atoms on the left and #2(2n+2)# on the right! #color(blue)(sqrt"")#

Thus, our equation is balanced in general.

Let's try it out! Here are two example solutions using the above general solution.

Methanol:

#2CH_3OH(l) + 3O_2(g) -> 2CO_2(g) + 4H_2O(l)#

Butanol:

#1/2 xx [2C_4H_9OH(l) + 12O_2(g) -> 8CO_2(g) + 10H_2O(l)]#

Verify that these are balanced.