Question

# When 0.109 g of Zn(s) combines with enough HCl to make 55.7 mL of HCl(aq) in...

When 0.109 g of Zn(s) combines with enough HCl to make 55.7 mL of HCl(aq) in a coffee cup calorimeter, all of the zinc reacts, which increases the temperature of the HCl solution from 23.2 °C to 24.8 °C: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Calculate the enthalpy change of the reaction ΔHrxn in J/mol. Insert your answer in kJ, but do not write kJ after the number. (Assume the density of the solution is 1.00 g/mL and the specific heat capacity of solution is 4.184 J/g°C.)

The amount of heat absorbed by solution , q = mcdt

Where

m = mass of solution = volume x density = 55.7 mL x 1.00 g/mL = 55.7 g

c = pecific heat capacity of solution = 4.184 J/g°C

dt = change in temperature = 24.8 - 23.2 = 1.6 oC

Plug the values we get q = 372.9 J

This amount of heat was liberated by 0.109 g of Zn

Thatmeans 0.109 g of Zn liberates 372.9 J

1 mol = 65.4 g of Zn liberates (65.4x372.9) / 0.109 = 223.7x103 J /mol

Therefore the enthalpy change of the reaction 223.7x103 J /mol

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