Question

The flame produced by the burner of a gas (propane) grill is a blue color when enough air mixes with the propane (C3H8) to burn it completely. For every gram of propane that flows through the burner, what volume of air is needed to burn it completely?

Answer 1

I got `"16.95 L"` of air (which is a lot!). Only about `"3.56 L"` of that air is `"O"_2`, though.

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Well, the main question is really, for `"1 g"` of `"C"_3"H"_8`, how many `"g"` of `"O"_2` do you need to react completely, and how many `"g"` of air you need to accomplish that many `"g"` of `"O"_2`?

First, write the reaction and balance it to get:

`"C"_3"H"_8(g) + 5"O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O"(g)`

(the water is a gas since it is open to the air and thus doesn't condense.)

With `"1 g"` of propane, you have

`cancel("1 g C"_3"H"_8) xx ("1 mol C"_3"H"_8)/(44.0962 cancel"g") = "0.0227 mols propane"`

Since the mol ratio of `"C"_3"H"_8:"O"_2` is `1:5`, you'll need `color(green)("0.1134 mols O"_2)` to react completely with this amount of propane.

Now, you were given the mol fraction of `"O"_2(g)` in the air as

`chi_(O_2(g)) = 0.210 -= (n_(O_2(g)))/(n_(t ot))`

The nice thing about mol fractions is that they are normalized to `1`, meaning all the mol fractions of everything in a mixture adds up to `1`.

So, if `n_(t ot) = 1.000`, then `n_(O_2(g)) = 0.210`. Consequently...

`0.210/1.000 = ("0.1134 mols O"_2)/("??? mols air")`

And to match up to this mol fraction, we need `color(green)"0.5399 mols"` of air. Finally, assuming air is an ideal gas, we can approximate the volume of air needed to completely burn `"1 g"` of propane:

`PV = nRT`,

`P` is in `"atm"` if `R` is in `"L"cdot"atm/mol"cdot"K"`. `V` is in `"L"` as a result. `n` is in `"mol"`s, and `T` is always in `"K"`. So, we get a volume of:

`=> bbV ~~ (nRT)/P`

`= (("0.5399 mols air")("0.082057 L"cdot"atm/mol"cdot"K")(205.0 + 273.1"5 K"))/("1.250 atm")`

`=` `ulbb("16.95 L")` of air