Question

When 0.113 g of Zn(s) combines with enough HCl to make 53.6 mL of HCl(aq) in a coffee cup calorimeter, all of the zinc reacts, which increases the temperature of the HCl solution from 23.3 °C to 24.7 °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.) Please show all work.

Answer 1

Volume of HCl = 53.6 ml

Density of solution = 1.00 g/ml

Mass of HCl = volume*density = 53.6*1.00 = 53.6 g

Mass of HCl = m = 53.6 g

Specific heat capacity of solution = c = 4.184 J/g ^oC

Temperature change = T = 24.7-23.3 = 1.4 ^oC

Heat change = Q = -mcT = 53.6*4.184*1.4 = -313.9 J

Mass of Zn = 0.113 g

Molar mass of Zn = 65.4 g/mol

Moles of Zn = mass/molar mass = 0.113/65.4 = 0.0017 mol

Enthalpy change of reaction = Q/moles of Zn = -313.9/0.0017 = -184686 J/mol = -184.7 kJ/mol

Enthalpy change of reaction = -184.7 kJ/mol