Question

How many liters of Propane gas (C3H8) will undergo complete combustion with 34 liters of Oxygen gas?

Answer 1

Well, let us write a stoichiometric equation to represent...(initially I had, `...."lets right"`...dohh...

Explanation:

`C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(g) + Delta`

Now at constant temperature and pressure, the which we assume, volume is proportional to the number of moles. And so if we use a `34*L` volume of `"dioxygen gas"` we will be requiring `(34*L)/5=6.8*L` with respect to `"propane gas...."`

Answer 2

The volume of propane that will react completely with `"34 L"` of oxygen gas is `color(blue)("6.8 L"`.

Explanation:

Balanced equation

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

Because we are working with gases, and we want to end up with liters of propane, we will be using the molar volume of a gas to convert between liters and moles, and moles and liters.

Since the conditions were not given, I'm going to use the current STP recommended by the IUPAC. The temperature for the current STP is `0^@"C"` `("273.15 K")` and the pressure is `10^5` `"Pa"` `("1 bar")`. Under these conditions, the molar volume of a gas is `"22.711 L/mol"`.

https://www.engineeringtoolbox.com/stp-standard-ntp-normal-air-d_772.html

https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure

Moles `"O"_2"`

Determine mol `"O"_2` from the molar volume at STP.

`34color(red)cancel(color(black)("L O"_2))xx(1"mol O"_2)/(22.711color(red)cancel(color(black)("L O"_2)))="1.5 mol O"_2"`

Moles `"C"_3"H"_8"`

Multiply mol `"O"_2` by the mole ratio between `"C"_3"H"_8"` and `"O"_2"` from the balanced equation, with `"C"_3"H"_8"` in the numerator.

`1.5color(red)cancel(color(black)("mol O"_2))xx(1"mol C"_3"H"_8)/(5color(red)cancel(color(black)("mol O"_2)))="0.30 mol C"_3"H"_8"`

Liters `"C"_3"H"_8"`

Multiply mol `"C"_3"H"_8"` by the molar volume at STP.

`0.30color(red)cancel(color(black)("mol C"_3"H"_8))xx(22.711"L C"_3"H"_8)/(1color(red)cancel(color(black)("mol C"_3"H"_8)))="6.8 L C"_3"H"_8"` (rounded to two significant figures)

The volume of propane that will react completely with `"34 L"` of oxygen gas is `"6.8 L"`.