1. Burning butane (C4H10) produces gaseous carbon dioxide and water.

[recoil33]

1. Burning butane (C4H10) produces gaseous carbon dioxide and water. The enthalpy of combustion of butane is -2650 kJ/mole. Determine how much water you can heat from room temperature (22 Degrees) to boiling with 1 kg of butane.

Therefore:

13O₂ + 2C₄H₁₀ ---Heat---> 8Co_{2 (g)} + 10H₂O_(l)

n(H2o) = 10 n(c4h10)

Enthalpy (c4h10) = -2650 kJ/mol
Temp (initial) = 22.0 degrees celcius
Mass (c4h10) = 1kg.
-------------------------------------------------------

n(c4h10) = 1000/58.12
= 17.20mol

(Although C4H10 would be a liquid therefore n = cV, wheres i do not know the concentration).

n(h20) = 5n(c4h6)
n(h20) = (17.20*5) * (18.016) =
= 1549g

m(h20) = 1.549kg

Now, to figure out how much the change in temperature will be. I assume use the equation q = mc(delta)T.


[Any advice please, don't reallly know where to start.
I should be able to figure this out, once i know where from.
]
Thanks, recoil33

 

[noleguy33]

I'm not following some of your notation, but here is what I would do-

Find out how many moles of Butane you have, which I think you did.

Next, use the enthalpy of combustion to find out the total amount of energy you can get from 1 Kg of butane. Next, use that value as your 'q' in your mc(delta)t equation and solve for m.

If your final answer needs to be a volume, use the density of water to convert.

I don't think this problem has anything to do with the water created by the combustion of Butane.

 

[recoil33]

Sorry, I didn't even read the question right -.-'

It's really simple, thanks anyways.

(I didn't realize that they said given 22-100 degrees, whereas i thought the change in temperature was unknown, as well as the mass.)