Question

When 1.34 g Zn(s) reacts with 60.0 mL of 0.750 M HCl(aq), 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:

Answer 1

Step-by-step explanation:

Zn + 2HCl = ZnCl₂ + H₂

1 mole 2 mole

1.34 g = 1.34 / 65. = .0206 mole

60 mL of .75 M HCl = .045 mole of HCl

so full .0206 mole of Zn will react with .0412 mole of HCl

.0206 mole of Zn gives out 3.14 kJ of heat

1 mole of Zn will react to give 152.43 kJ of heat

enthalpy change of heat per mole of Zn

= 152.43 kJ .