2C2H6+702---->4CO2+6H20

DELTAH= -3120.6 kJ/mol
What mass of ethane must be combusted in order to heat 366 g of water from 25 degree celcius to 100 degree celcius? Assume the heat transfer is 100% efficient.
a)44.1 g
b) .903g
c) 3.61g
d) 2.21 g
Should I approach this problem by q=m*s*delta T??
I know the answer is d. I don't know how to approach the problem

3 answers

Yes, that approach will get you there, at least that's a good place to start.
How much heat do you need to heed the water? That's
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) = ?
Then 3120.6 kJ/mol x (grams ethanol/2*(molar mass C2H6) = q
Solve for grams ethanol and round to 3 significant figures.
I have got the answer but can you explain me which formula you used after calculating q in part 1?
The easy way is to say I used good old fashioned reasoning but that's not the kind of answer you're looking for.
I think q comes out to be approx 115 kJ; let's just say that that's the answer for q = mcdT.
Then here is what we have.
The molar mass of ethane (sorry--I wrote ethanol in the response--my bad) is 30 so 2*30 = 60 g ethane produced 3120.6 kJ. We want to know know how many g will produce 114.8 kJ. We set up a proportion like this
(60g/3120.6 kJ) = (x g/114.8) so
x = 60*114.8/3120 = 2.207 = 2.21 g.
Instead of setting up a proportion I cobbled together a linear expression that basically says
known kJ heat by ethane/g x grams = heat needed.
Hope this helps.
By the way, I think there is an error in the problem. When it says dH is -3120.6 kJ/mol they really mean -3120 kJ/rxn (thats for 2 mols). It 3120.6/2 kJ/mol.
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