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

##### 2 Answers

#### Answer:

Balance mass and balance charge........

#### Explanation:

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

Odd-numbered alcohols require non-integral stoichiometric coefficients.....you can always remove the

#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

Each alcohol follows the same general formula,

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)#

**Balance the carbons**

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

We now have

**Balance the hydrogens**

Given

#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

**Balance the oxygens**

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

The

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

Now we have

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

#=> x = 3/2n#

Plug in

#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.