Question

If air is `20.9%` oxygen by volume, 1. How many liters of air at `25.5^@ "C"` and `1.000` atm are needed for the complete combustion of `15.0 "kg"` of propane vapor? 2. What volume of each product is produced?

Thanks in advance

Answer 1

`V("air")=1.99*10^2color(white)(l)"L"`

`V("CO"_2)=25.0color(white)(l)"L"`
`V("H"_2"O")=33.3color(white)(l)"L"`

Explanation:

Volume of air consumed

Starting by writing a balanced equation for the combustion of propane in air.

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

The combustion of each mole of propane would consume five moles of oxygen. Propane has a formula mass of

`3*12.01+8*1.008=44.09`

`15.0color(white)(l)"kg"` of propane contains

`(15.0*10^3color(white)(l)color(red)(cancel(color(black)("g"))))/(44.09color(white)(l)color(red)(cancel(color(black)("g")))*"mol"^(-1))=3.40*10^2color(white)(l)"mol"`

... molecules. Complete combustion would consume

`3.40*10^2color(white)(l)"mol"color(red)(cancel(color(black)("C"_3"H"_8)))*(5"O"_2)/(color(red)(cancel(color(black)("C"_3"H"_8))))=1.70*10^3color(white)(l)"mol O"_2`

`1.000color(white)(l)"atm"=1.013*10^5color(white)(l)"Pa"`

`T=273.15+25.5=298.6color(white)(l)"K"`

Apply the ideal gas law,

`V_("O"_2)=(color(grey)(n("O"_2))*R*T)/(P)=(color(grey)(1.70*10^3)*8.31*298.6)/(1.013*10^5)=41.6color(white)(l)"L"`

According to the question,

`V("O"_2)=color(green)(20.9%)*V_"air"`

hence

`V("air")=(V("O"_2))/(color(green)(20.9%))=1.99*10^2color(white)(l)"L"`

Volume of each product

Again, referring to the balanced equation:

`n("CO"_2)=3*n("C"_3"H"_8)=3*3.40*10^(2)=1.02*10^3color(white)(l)"mol"`

`n("H"_2"O")=4*n("C"_3"H"_8)=4*3.40*10^(2)=1.36*10^3color(white)(l)"mol"`

Apply the ideal gas law:

`V("CO"_2)=(color(grey)(n("CO"_2))*R*T)/(P)=(color(grey)(1.02*10^3)*8.31*298.6)/(1.013*10^5)=25.0color(white)(l)"L"`

`V("H"_2"O")=(color(grey)(n("H"_2"O"))*R*T)/(P)=(color(grey)(1.36*10^3)*8.31*298.6)/(1.013*10^5)=33.3color(white)(l)"L"`

Alternatively, given that

`n_1/V_1=n_2/V_2`

... for two ideal gases under the same temperature and pressure, after calculating the volume of any gaseous species involved it would be possible to derive the volume of the rest (of the gaseous products) straight from their molar ratio without having to apply the ideal gas law. That is:

`V("CO"_2)=color(blue)(V("O"_2))*(n("CO"_2))/(n("O"_2))`

`V("H"_2"O")=color(blue)(V("O"_2))*(n("H"_2"O"))/(n("O"_2))`

Try evaluating these two expressions yourself and see if you arrive at the same results.