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 o

Question

Natasha2012

17 days ago

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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 o

f zinc reacting for the reaction:Zn(s) + 2HCl(aq) --> ZnCl₂(aq) + H₂(g)

Chemistry

1 answer:

Alekssandra [2.7K] · Answer 1 · 2026-03-19T20:46:14+03:00

Answer: The change in enthalpy for each mole of zinc involved in the reaction is 152.4 kJ/mol.

Explanation:

First, we need to determine the moles of Zn and HCl.

$\text{Moles of }Zn=\frac{\text{Mass of }Zn}{\text{Molar mass of }Zn}$

The molar mass of Zn is 65 g/mole

$\text{Moles of }Zn=\frac{1.34g}{65g/mole}=0.0206mole$

and,

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}=0.750M\times 0.0600=0.0450mole$

Next, we must identify the limiting reagent and the excess reagent.

The chemical reaction given is:

$Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$

According to the balanced reaction we find that

1 mole of $Zn$ reacts with 2 moles of $HCl$

Thus, 0.0206 moles of $Zn$ react with $0.0206\times 2=0.0412$ moles of $HCl$

This leads us to determine that $HCl$ is the excess reagent because the moles provided exceed the required moles, while $Zn$ is limiting and restricts product formation.

Now to find the enthalpy change for each mole of zinc reacting in this reaction.

From the reaction we gather that,[ [TAG_59]]

0.0206 moles of Zn yield heat = 3.14 kJ

This implies that 1 mole of Zn generates heat = $\frac{3.14kJ}{0.0206mol}=152.4kJ/mol$

Hence, the enthalpy change per mole of zinc involved in this reaction amounts to 152.4 kJ/mol.