[Rutherford]

15 ml of a gaseous hydrocarbon are mixed with 120 ml oxygen and ignited. After the reaction, the burned gases are shaken with concentrated aqueous KOH solution. A part of the gases is completely absorbed while 67.5 ml gases remain. It has the same temperature and pressure as the original unburned mixture. What is the chemical formula of the hydrocarbon used for the experiment?

15 C_xH_y + 15(x+y/4) O₂ 15x CO₂ + 15y/2 H₂O
The starting volume was 135 ml. The decrease in volume is (assuming water is in gas state): 15+15x+15y/4+15x-15y/2, so i have:
135 - (15+15x+15y/4+15x-15y/2) = 67.5
For y=2, x=2, therefore it is acetylene, but in the answer it is ethane. Did I made a mistake somewhere, or is their answer wrong?

[sjb]

15 ml of a gaseous hydrocarbon are mixed with 120 ml oxygen and ignited. After the reaction, the burned gases are shaken with concentrated aqueous KOH solution. A part of the gases is completely absorbed while 67.5 ml gases remain. It has the same temperature and pressure as the original unburned mixture. What is the chemical formula of the hydrocarbon used for the experiment?

15 C_xH_y + 15(x+y/4) O₂ 15x CO₂ + 15y/2 H₂O
The starting volume was 135 ml. The decrease in volume is (assuming water is in gas state): 15+15x+15y/4+15x-15y/2, so i have:
135 - (15+15x+15y/4+15x-15y/2) = 67.5
For y=2, x=2, therefore it is acetylene, but in the answer it is ethane. Did I made a mistake somewhere, or is their answer wrong?

Are you sure that the oxygen was not in excess?

[Dan]

I also get ethane.

1. Write an expression for the change in volume due to combustion alone (before KOH treatment).
2. Write an equation for the change in volume due to KOH treatment (after combustion).
3. The total volume change is the sum of 1. and 2.

[Rutherford]

Combustion alone:
135-15-15x-15y/4+15y/2=z
z-15x=67.5

135-15-15x-15y/4+15y/2-15x=67.5 same equation as in my original post. I don't understand  .

[Dan]

What is z?

The change in volume due to combustion = -(volume of hyrocarbon)-(volume of reacting oxygen)+(volume of CO₂ and water)
The change in volume due to KOH treatment = -(volume of CO₂ and water)

Therefore:

The overall change in volume = -(volume of hyrocarbon)-(volume of reacting oxygen)+(volume of CO₂ and water)-(volume of CO₂ and water) = 67.5-135 mL

[Rutherford]

Okay, saw now where was my mistake. I wrote that CO₂ volume was only lost, but I didn't write that it was firstly produced, so it cancels out. For water, I wrote the opposite. It is ethane then. Thanks for the help.

[Needaask]

I assumed that the H2O would liquify. So the change in volume=67.5-135=-67.5=-15-15x-15y/4
And since the carbon dioxide and water adds up to 67.5, 15x+15y/2=67.5

So equation (1): 15x+15y/4=52.5
and eqn (2): 15x+15y/2=67.5

so y=4 and x=2.5 what went wrong for me here?

[Dan]

the carbon dioxide and water adds up to 67.5

Why/how?

[Needaask]

the carbon dioxide and water adds up to 67.5

Why/how?

Sorry it shouldn't be 67.5ml I forgot about the mole ratio and volume of gases rule. But then now I can't think of another equation with x and y to solve the simultaneous equation with..

Thanks for the help

[Dan]

But then now I can't think of another equation with x and y to solve the simultaneous equation with..

Neither can I, but eq 1:

15x + 15y/4 = 52.5

4x + y = 14

By inspection there are 3 solutions where x > 0 < y:

x = 1; y = 10
x = 2; y = 6
x = 3; y = 2

The only solution that makes chemical sense is x = 2; y = 6 (ethane)

[Needaask]

But then now I can't think of another equation with x and y to solve the simultaneous equation with..

Neither can I, but eq 1:

15x + 15y/4 = 52.5

4x + y = 14

By inspection there are 3 solutions where x > 0 < y:

x = 1; y = 10
x = 2; y = 6
x = 3; y = 2

The only solution that makes chemical sense is x = 2; y = 6 (ethane)

Oh we are supposed to find that out this way? I didn't expect that.

But actually when doing this we assumed the excess reagent was the oxygen, but shouldn't we be able to assume that the hydrocarbon is the excess reagent too?

[Dan]

If O₂ was the limiting reagent, then all the O₂ would be consumed - i.e. the change in volume would have to be >120 mL. The fact that this is not true unambiguously indicates that O₂ is in excess.

[Needaask]

If O₂ was the limiting reagent, then all the O₂ would be consumed - i.e. the change in volume would have to be >120 mL. The fact that this is not true unambiguously indicates that O₂ is in excess.

Oh that's right! If it was the excess all of it would be used up and only some CxHy would be left behind. So the remaining volume has to be >15ml?