Chemistry help; calorimeter, Calculate delta H?

Question

(thanks bonhommecretienne...but need a more complete answer)

A 100mL sample of 0.200 M aqueous hydrochloric acid is added to 100 mL 0f 0.200 M aqueous ammonia in a calorimeter whose heat capacity is 480 J/degreesC. The following reaction occurs when the two solutions are mixed.

HCl (aq) + NH3(aq) ---------> NH4Cl (aq)

The temperature increase is 2.34 degreesC. Calculate deltaH of neutralization per mole of HCl and NH3 reacted. Assume the densities and the specific heat are the same as for the water. (1.00 g/mL and 4.184 J/g degreesC respectively)

Answer 1

Heat released = 480*2.34 J

So 0.02 mol releases this much

1 mol will release 480*2.34/0.02 = 56.16 kJ

Answer 2

a million.) 681.2 kj/mol 2.) 80.3 J/mol 3.) 665.14 kj/molK all 3 are the sum of issues minus the sum of reactants. yet via fact which you're no longer given values for gibbs loose potential you will discover it with the equation deltaG= deltaH- temp x deltaS. As for the sencond a million/2 of each question I forget approximately approximately and that i dont desire to respond to incorrectly