Question

When 0.500 g CaO(s) is added to 140. g H2O at 23.1 °C in a coffee...

When 0.500 g CaO(s) is added to 140. g H2O at 23.1 °C in a coffee cup calorimeter, this reaction occurs.

CaO(s)+H2O(l)-->Ca(OH)2 (aq) and enthalpy = -81.9 kj/mol

Calculate the final temperature of the solution. The specific heat capacity of water is 4.184

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Answer #1

Hrxn = -81.9 kJ/mol

mol of CaO = mass/MW = 0.5/56.0774 = 0.0089162479 mol of CaO

mol of H2O = mass/MW = 140/18 = 7.777 mol of H2O

since CaO is limiting

then

0.0089162479 mol should be used

Q = nHRxn = 0.0089162479 *-81.9 = -0.73024070301kJ = -730.240 kJ

note that this is EXOthermic, meaning this will increase the temperature of the system (i.e.calorimeter)

then

Q = m*Cp*dT

Q = 730.240 J

mT = 0.5+140 = 140.5 g

Cp = 4.184J/gC (approx of water)

dT = Tf- 23.1

Q = m*Cp*dT = (140.5)(4.184)(Tf-23.1) = 730.240

(140.5)(4.184)(Tf-23.1) = 730.240

Tf = 730.240 /(140.5*4.184) + 23.1 = 24.3422

Tf = 24.3422 °C

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