Question #59287
Calculate the number of Btu evolved at 25 °C when 80 lb pf ZnCl2 are added to 200 lb of water?
Expert's answer
Btu – British thermal unit
1 Btu = 1055.06 J = 1.06 kJ
1 lb = 0.454 kg
80 lb = 36.32 kg
200 lb = 90.8 kg
M (ZnCl2) = 136 g/mol
M (H2O) = 18 g/mol
n (ZnCl2) = 36 320/136 = 267 mol
n (H2O) = 90 800/18 = 5 044 mol
Enthalpy of dissolution of ZnCl2 at this temperature and at this proportion to water is -66 kJ/mol
Therefore the amount of heat emitted is: 66*267 = 17 589 kJ =16 593 Btu
1 Btu = 1055.06 J = 1.06 kJ
1 lb = 0.454 kg
80 lb = 36.32 kg
200 lb = 90.8 kg
M (ZnCl2) = 136 g/mol
M (H2O) = 18 g/mol
n (ZnCl2) = 36 320/136 = 267 mol
n (H2O) = 90 800/18 = 5 044 mol
Enthalpy of dissolution of ZnCl2 at this temperature and at this proportion to water is -66 kJ/mol
Therefore the amount of heat emitted is: 66*267 = 17 589 kJ =16 593 Btu
