# The combustion of a gaseous alkane produces 7 L of carbon dioxide plus water vapour. If the original volume of alkane plus oxygen was 6 L, what is the formula of the alkane?

Jun 2, 2017

${C}_{3} {H}_{8}$

#### Explanation:

The reaction used is a normal total oxidation of the hydrocarbon by oxygen:

${C}_{n} {H}_{\text{2n+2}} + \left[n + \left(\frac{n + 1}{2}\right)\right] {O}_{2} \to n C {o}_{2} + \left[n + 1\right] {H}_{2} O$

The coefficient for the O_2 is coming from the addition of the ones for the products.

Using the ideal gas formula PV=nRT, where everything is constant but V and n, we can relate both states, before and after the reaction.

${V}_{1} / {V}_{2} = {n}_{1} / {n}_{2} = \frac{600}{700} = \frac{6}{7}$

where ${n}_{1}$ and ${n}_{2}$ are the total number of moles, for the first side of the equation that means the oxygen and the hydrocarbon, while in the second the carbon dioxide and the water.

Then the first side

${n}_{1} = 1 + \left[n + \left(\frac{n + 1}{2}\right)\right] = 1 + \left(\frac{3 n + 1}{2}\right) = \frac{3 n + 3}{2} = 6$

solving n

$\frac{3 n + 3}{2} = 6$

$3 n + 3 = 12$ factorizing 3

$n + 1 = 4$

$n = 3$

Then the formula is ${C}_{3} {H}_{8}$

Now, the same result should be obtained from the other side, lets try it out.

${n}_{2} = n + \left(n + 1\right) = 2 n + 1 = 7$

$2 n + 1 = 7$

$n = \frac{7 - 1}{2}$

$n = 3$