Calorimetry questions

I have a couple of questions on calorimetry that I'm confused about and I was wondering if anyone can please tell me where I'm going wrong. Sorry if my explanations seem too pedantic. :s-smilie:

The first question involves 25.0cm^3 of 1.0moldm^-3 hydrochloric acid being added to 25.0cm^3 of 1.0moldm^-3 sodium hydroxide in a polystyrene calorimeter. I'm unsure as to whether I need to use just the mass of the sodium hydroxide or the mass of both the aqueous solutions. The question also says that 0.025mol of acid and base reacted to form 0.025mol of water. The equation I'm given is:

HCl(aq) + NaOH(aq) --> NaCl(aq) + H_2 O(l)

So far I have:
Q=25 x 4.2 x (19.5-26.0)
=-682.5
deltaH =

\frac{-682.5}{0.025}

=-27300J

=27.3kJmol^{-1}

The second question is:
A spirit burner containing ethanol was weighed and found to be 56.60g. It was ignited and used to heat up a copper calorimeter (of negligible heat capacity) containing 200cm^3 of water at 22.0 degrees C. When the temperature of the water was 38.0 degrees C, the burner was blown out and reweighed and found to be 55.68g. Calculate the enthalpy of combustion of ethanol.
At the end of the answer page it says I have to compare the obtained value with the value in the data book, which is -1370kJmol^-1. So far I have:
56.60g - 55.68g = 0.92g
deltaT = 22.0 - 38.0 = 16.0
Q = 200 x 4.2 x -16.0
=-13440
deltaH =

\frac{-13440}{n}

n=\frac{200}{18}=11.11mol

deltaH =

\frac{-13440}{11.11}

=-1210J
=

-1.21kJmol^{-1}

Then I tried dividing 0.92g by the molar mass of ethanol for n, so that

\frac{-13440}{0.02}

-672kJmol^{-1}

OK, so this is the closest answer I have to -1370, but I don't think I could say a difference that big is due to experimental error. :s-smilie:

The next question also involves working out the enthalpy change of the combustion of ethanol:
0.046g of ethanol was burned in an improved flame calorimeter and produced a temperature rise of 4.0 degrees C. An electrical heater connected to a joulemeter required 1310J to produce the same temperature rise in the apparatus.
My answer so far:
Q = 0.046 x 4.2 x -4 = 0.7728

n= \frac{0.046}{46}

\frac{0.7728}{0.001}=772.8

\frac{0.046}{4}

= 0.0115

\frac{1310}{0.0115}

= 113913
Q = 0.046 x 113913 x -4 = -20960
deltaH =

\frac{-20960}{0.001}

= -20960000
=

-21000kJmol^{-1}

Which isn't anywhere near -1370 :confused:

So I tried:

\frac{1310}{4}=327.5

Q = 0.046 x 327.5 x -4 =-60.26
deltaH =

\frac{-60.26}{0.001}

=-60260
=

-60kJmol^{-1}

I've probably got something really trivial wrong, but I've got such a wide range of answers that aren't anywhere near what the enthalpy of the combustion of ethanol is supposed to be and I'm really confused! Apologies for the long post but I didn't want anyone thinking I was being lazy and getting someone else to do my homework. Thanks in advance for any advice. :smile: