Chemistry
posted by David on .
How many grams of butane must be burned to provide the heat needed to melt a 85.0g piece of ice (at its melting point) and bring the resulting water to a boil (with no vapourization)? Assume a constant pressure of 1 bar and use the following data:
Hfo(H2O(g))=241.83 kJ/mol, Hfo(O2(g))=0 kJ/mol, Hfo(CO2(g))=393.5 kJ/mol, Hfo(C4H10(g))=124.7 kJ/mol
Tfus(H2O)=0 oC, Tvap(H2O)=100 oC
Hfus(H2O)=6.01 kJ/mol, Cp(H2O(l))=75.291 J/mol/oC

2C4H10 + 13O2 ==> 8CO2 + 10H2O
Use the Hf information to find delta H (DH) for the reaction .
DHrxn = (n*DHf products)  (n*DHf reactants). That is the number of joules (or kJ but keep it all straight) available from the reaction as written.
Now, how much heat do you need.
You need to melt the ice.
mass ice x heat fusion = ?
You need to raise T of H2O from zero C to 100 C.
mass ice melt (85g) x specific heat water x 100 = ?
The sum of those two = amount of heat needed.
g C4H10 needed = 116g x (kJ needed/kJ rxn)
Check my work. It's late here. 
Okay, so I got:
DHrxn = 8(395.5)+10(241.83)[2(127.7)+13(o)
=5326.9 kJ
mass(moles) ice x heat fusion 4.718mol(85/18.016)*6.01KJ/mol? =28.355Kj
Then I have 4.718mol *75.291J/mol/oC x 100= 35522.2938J
so sum together= 28.355kJ + 35.522 =
63.877kJ
then 2(58) x (63.877/53269kJ))=0.1391g
I'm not sure if I did this right I may have went wrong in the first step, so if you could steer me in the right direction it would be appreciated,
Thanks. 
I found one SMALL math error and one large math error.
You used DHfCO2 = 395.5 but your post (and my text) shows 393.5 but that makes only a small difference in the problem. . The larger error is the last step where you typed in 53269 and you omitted the decimal; it should be 5326.9(actually with the correction of the CO2 I have 5316.9 for this). That changes the answer by a factor of 10 so it is 1.39 g C4H10 needed. 
How did you get g C4H10 needed??? I'm also working on this question

g C4H10 needed = 2(58) x (63.877/5316.9) = ?

thanks!