# Propane is combusted completely in air. What volume of dioxygen gas is required to combust a 200*mL volume of propane?

Apr 19, 2017

Well, the stoichiometric equation is............

#### Explanation:

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(g\right)$

Is this balanced with respect to mass and charge? If it is not, then we cannot use it to represent reality. The equation tell you that one equiv of propane reacts with five equiv dioxygen to give three equiv carbon dioxide, and four equiv water (of course the water may condense!, but this is our assumption). This relationship goes back to old Avogadro..........$\text{equal volumes of gases}$ contain $\text{equal NUMBERS of molecules}$.

And so, given the stoichiometry, we need 5 volumes of dioxygen to 1 volume of propane.......and so for your question, for every $1 \cdot L$ of ${O}_{2} \left(g\right)$ we need a $200 \cdot m L$ volume of $\text{propane}$. Are you with me?

What about methane gas? Certainly, this is a common fuel, and could be cooking your breakfast right now.