# What us the molar enthalpy of Ba(SCN)2 (s) ? In a calorimeter 10.0 g of Ba(OH)2 8H2O (s) is mixed with 10.0 g NH4SCN (s). the equation is:  Ba(OH)2 8H2O (s) + 2NH4SCN (s) -> Ba(SCN)2 (s) + 10H2O(l) +2NH3 (g) The temperture starts at 25 celcius and drops to 4.56 celcius. The specifec heat capacity of the 1.560 L of water in the calorimeter is 4.187 J/g celcius. Calculate  the molar enthalpy of Ba(SCB)2 (s) produced.

To calculate the molar enthalpy of Ba(SCN)2 first we have to get the moles of the BA(SCN)2 from the solution using stoichiometry.

1. Check if the equation is balanced. The equation is balanced.

2. Get the moles of Ba(SCN)2. Apply the limiting reagent method.

Ba(OH)2*8H2O + 2 NH4SCN = Ba(SCN)2 + 10 H2O + 2 NH3

10 g Ba(OH)2*8H2O x ( 1mole Ba(OH)2*8H2O/ 315.4650g Ba(OH)2*8H2O) x ( 1mole Ba(SCN)2 /1mole Ba(OH)2*8H2O)

= 0.0317 moles Ba(SCN)2 ---> limiting reagent

10 g NH4SCN x ( 1mole NH4SCN/ 76.1215 grams) x (1mole Ba(SCN)2 / 2 moles NH4SCN)

= 0.06568 moles Ba(SCN)2

2. We get the heat absorbed by the water.

Q = mC*deltaT

= 1560 g (4.187 J/ gC) (277.56 - 298)

= -133508 J = -133.5 kJ

delta H = -Q

delta H = - (-133.5) = 133.5 kJ

molar enthalpy of Ba(SCN)2 = deltaH / moles Ba(SCN)2

= 133.5/0.0317moles

= 4211kJ/mole --> final answer :)

notes:

-we converted the temperature to kelvin

-the moles of Ba(SCN)2 used is the one derived from the limiting reagent

-delta T = (Tfinal - T initial)

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