Question

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

Answer 1

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.

Answer 2

`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`.

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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.