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recoil33
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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---> 8Co2 (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
Therefore:
13O2 + 2C4H10 ---Heat---> 8Co2 (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
Last edited: