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ChemistryPhysical Chemistry

Enthalpy Change Formula: Find ΔH per Mole of Zinc

Zinc is limiting because 0.0450 mol HCl exceeds the 0.0410 mol required. Dividing the released heat, −3.14 kJ, by 0.0205 mol Zn gives ΔH = −153 kJ mol⁻¹ Zn.

Question

A 1.34 g1.34\,\mathrm g sample of Zn(s)\mathrm{Zn(s)} reacts with 60.0 mL60.0\,\mathrm{mL} of 0.750 M HCl(aq)0.750\,\mathrm M\ HCl(aq) according to

Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)\mathrm{Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)}

and releases 3.14 kJ3.14\,\mathrm{kJ}. Determine ΔH\Delta H per mole of Zn consumed.

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Evidence boundary

This result assumes constant-pressure conditions and assigns the measured 3.14 kJ entirely to the stated reaction. It verifies Zn as the limiting reactant and reports the value per mole of Zn, as requested. The negative sign denotes heat released; no calorimeter or environmental heat-loss correction is available from the prompt.