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

2 Answers
May 12, 2017

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....

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.

May 12, 2017

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.