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:

\({\bf{Zn(s) + 2HCl(aq)}} \to {\bf{ZnC}}{{\bf{l}}_{\bf{2}}}{\bf{(aq) + }}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

Short answer

Enthalpy change = -153.86 kJ.

Step-by-step solution

Number of moles

The given reaction is:

\({\rm{Zn(s) + 2HCl(aq)}} \to {\rm{ZnC}}{{\rm{l}}_{\rm{2}}}{\rm{(aq) + }}{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\)

The reaction used one mole of zinc. The number of moles is evaluated as:

\({\bf{Number of moles = }}\frac{{{\bf{Given mass}}}}{{{\bf{Molar mass}}}}\)

\(\begin{array}{c}{\rm{Number of moles = }}\frac{{{\rm{1}}{\rm{.34g}}}}{{{\rm{65}}{\rm{.38g/mol}}}}\\{\rm{ = 0}}{\rm{.02 mol}}\end{array}\)

So, 0.02 mol of zinc produces - 3.14 kJ heat.

Calculation of enthalpy change

0.02 mol of zinc produces 3.14 kJ heat.

So, one mole of zinc will produce = \(\frac{{{\rm{3}}{\rm{.14}}}}{{{\rm{0}}{\rm{.02}}}}{\rm{ = - 157 kJ}}\)

\(\begin{array}{*{20}{l}}{{\rm{Enthalpy change = - 157 kJ + 3}}{\rm{.14 kJ}}}\\{{\rm{\;\;\;\; = - 153}}{\rm{. 86 kJ}}}\end{array}\)