# Combustion of which gas gives 7 moles of gaseous product from 6 moles of gaseous reactant?

## $1.$ CH_4; $2.$ H_2C=CH_2; $3.$ HC-=CH; $4.$ ${H}_{3} C - C {H}_{2} C {H}_{2}$

Jun 26, 2016

Propane combustion; $\text{option 4}$

#### Explanation:

${\underbrace{{C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right)}}_{\text{6 moles" rarr underbrace(3CO_2(g) +4 H_2O(g))_"7 moles}}$

Because $V \propto n \text{ (T and P constant)}$, the VOLUMES represent the stoichiometric equivalents of hydrocarbon and oxidant.

You should write out all the combustion reactions with each fuel to satisfy yourself that this is the case:

i.e. ${C}_{2} {H}_{4} \left(g\right) + 3 {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$, which stoichiometry disagrees with the experimental result.

You can do the rest.