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?

Question

Mateo Aguilar · Accepted Answer

#C_3H_8#

The hydrocarbon undergoes a typical total oxidation by oxygen in this reaction:

#C_nH_"2n+2"+[n+((n+1)/2)]O_2 -> nCo_2 + [n+1]H_2O#

The sum of the coefficients for the products yields the O_2 coefficient.

We can relate both states, before and after the reaction, using the ideal gas formula PV=nRT, where all other variables are constant except V and n.

#V_1/V_2 = n_1/ n_2 = 600/700 = 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.

Next, the initial side

#n_1 = 1 + [n+((n+1)/2)] = 1 +((3n+1)/2) = (3n+3)/2 = 6#

figuring out n

#(3n+3)/2 = 6#

#3n+3 = 12# factorizing 3

#n+1 = 4#

#n = 3#

Then the formula is #C_3H_8#

Let's test it out now; the opposite side ought to produce the same outcome.

#n_2 = n+(n+1) = 2n+1 = 7#

#2n+1 = 7#

#n = (7-1)/2#

#n = 3#

Kai Bowman · Answer

undefined Based on the given information, the balanced chemical equation for the combustion of the alkane is:

2CnH2n+2 + (3n+1)O2 → 2nCO2 + (n+1)H2O

Since the volume of carbon dioxide produced is 7 L, and the original volume of the alkane plus oxygen was 6 L, the number of moles of carbon dioxide produced is equal to the number of moles of alkane consumed.

Therefore, the alkane has the formula C2H6.