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

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....

${C}_{5} {H}_{11} O H \left(s\right) + \frac{15}{2} {O}_{2} \left(g\right) \rightarrow 5 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(l\right) + \Delta$

Odd-numbered alcohols require non-integral stoichiometric coefficients.....you can always remove the $\frac{1}{2}$ coefficient by doubling the equation.

May 12, 2017

$\textcolor{g r e e n}{2} {C}_{n} {H}_{2 n + 1} O H \left(l\right) + \textcolor{g r e e n}{3 n} {O}_{2} \left(g\right) \to \textcolor{g r e e n}{2 n} C {O}_{2} \left(g\right) + \textcolor{g r e e n}{\left(2 n + 2\right)} {H}_{2} O \left(l\right)$

$n = 1 , 2 , 3 . . .$ 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, ${C}_{n} {H}_{2 n + 1} O H$, or ${C}_{n} {H}_{2 n + 2} O$, where $n = 1 , 2 , 3 , . . .$.

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

${C}_{n} {H}_{2 n + 2} O \left(l\right) + {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + {H}_{2} O \left(l\right)$

1) Balance the carbons

${C}_{n} {H}_{2 n + 2} O \left(l\right) + {O}_{2} \left(g\right) \to n C {O}_{2} \left(g\right) + {H}_{2} O \left(l\right)$

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

2) Balance the hydrogens

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

$2 n + 2 = 2 m$,

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

${C}_{n} {H}_{2 n + 2} O \left(l\right) + {O}_{2} \left(g\right) \to n C {O}_{2} \left(g\right) + \left(n + 1\right) {H}_{2} O \left(l\right)$

We now have $2 n + 2$ 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, $x \ne n$.

$2 {C}_{n} {H}_{2 n + 2} O \left(l\right) + 2 x {O}_{2} \left(g\right) \to 2 n C {O}_{2} \left(g\right) + \left(2 n + 2\right) {H}_{2} O \left(l\right)$

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

$2 + 2 \left(2 x\right) = 2 \left(2 n\right) + \left(2 n + 2\right)$

$\implies x = \frac{3}{2} n$

Plug in $x$ to get:

$\textcolor{b l u e}{2 {C}_{n} {H}_{2 n + 2} O \left(l\right) + 3 n {O}_{2} \left(g\right) \to 2 n C {O}_{2} \left(g\right) + \left(2 n + 2\right) {H}_{2} O \left(l\right)}$

Now as a check, we have:

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

$2 C {H}_{3} O H \left(l\right) + 3 {O}_{2} \left(g\right) \to 2 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right)$

Butanol:

$\frac{1}{2} \times \left[2 {C}_{4} {H}_{9} O H \left(l\right) + 12 {O}_{2} \left(g\right) \to 8 C {O}_{2} \left(g\right) + 10 {H}_{2} O \left(l\right)\right]$

Verify that these are balanced.