Question

95 mL of 0.225 M AgNO3 was mixed with 47.5 mL of 0.225 M CaCl2 in a coffee cup calorimeter. If a reaction occured was it exothermic or endothermic? If the reacion started at 23.7 degrees Celsuis what is the final temperature of the solution?

Answer 1

2AgNO3 + CaCl2 = 2AgCl + Ca(NO3)2

mmol of Ag = MV = 95*0.225 = 21.375

mmol of Cl- = MV = 2*47.5*0.225 = 21.275

We need HRXN

HRxn = AgCl - (Ag+ + Cl-)

HRxn = (-127.0) - (105.8-167.2) = -65.6 k/mol

this is eoxthermic, since it will realease heat

for

mol of Ag = 21.275*10^-3

Qrxn = n*HJRXN = -65.6 * 21.275 /1000 = -1.39564 kJ = -1395.64 J

now...

mass of solution = 95+47.5 = 142.5 mL

Q = m*C*(Tf-Ti)

1395.64 = 142.5*4.184*(Tf-23.7)

Tf = 1395.64 /(142.5*4.184) + 23.7

Tf = 26.04°C