[Wsx594]

So I have 1.17g of CaCl2 which I add to 49.24 g H2O(l). When added I got the TΔ to be 3.5°C. After that I found the enthalpy, q. Which I got to be 50.41x 4.18x 3.5=q , q=-737.49 KJ. Now I found that 1.17g of CaCl2 was .0105 mol. Next I setup the equation 1/.0105 = 95.23 then I multiplied -737.49 x 95.23 to get the molar enthalpy of -70237.14 KJ/Mol which I then divided by the amount of grams to get -632.88 KJ per gram.
The second reaction I added 1g NH4NO3 to 49.01 g H2O(l) which gave me a temp change of TΔ of 1.3°C
 I then calculated enthalpy: q= 50.01x4.18 x 1.3, q=271.75
I Then found that 1 g of NH4NO3 was .0125 mol. I then divided 1/.0125= 80 which I then multiplied by 271.75 to get 21740 KJ/Mol
I took that number and divided it by 80 which gave me 271.75 KJ/g
Lastly I needed to find the amount of NH4NO3 that would cancel out a 10g sample of CaCl2 added to water.
So I multiplied 10(-632.88) to get -6328.8+ 271.75x=0
I then solved for X to get 23.29 g of NH4NO3 needed to balance out the endo and exo reactions. Where did I go wrong in my work?