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:
Zn(s) +2HCl (aq) → ZnCl2 (aq) +H2 (g)

Answer 1

Answer:

Approximately $\left(-1.53 \times 10^{2}\; \rm kJ\right)$ per mole $\rm Zn$. (The negative value suggests the release of heat.) Assumption: the reaction was complete.

Explanation:

Finding the limiting reactant

This question specified the quantity of both $\rm Zn$ and $\rm HCl$. Hence, either of the two species could be the limiting reactant. Calculating the number of moles of $\rm Zn\!$ that was actually consumed requires finding the limiting reactant.

Look up the relative atomic mass of $\rm Zn$ on a modern periodic table:

• $\rm Zn$: $65.38$.

In other words, $M(\mathrm{Zn}) = 65.38\; \rm g \cdot mol^{-1}$.

Calculate the number of moles of atoms present in that $1.34\; \rm g$ of $\rm Zn$:

$\begin{aligned}&n(\mathrm{Zn}) \\ &= \frac{m(\mathrm{Zn})}{M(\mathrm{Zn})} \\ &= \frac{1.34\; \rm g}{65.38\; \rm g \cdot mol^{-1}}\approx 0.020496\; \rm mol\end{aligned}$.

On the other hand, calculate the number of moles of $\rm HCl$ molecules in that $60.0\; \rm mL$ of $0.750\; \rm mol \cdot L^{-1}$ ($0.750\; \rm M$) ${\rm HCl}\, (aq)$ solution. Start by converting the unit of volume to liters (so as to match the unit of the concentration of this solution.)

$\begin{aligned}&V({\rm HCl})\\ & = 60.0\; \rm mL \times \frac{1\; \rm L}{10^{3}\; \rm mL} \\ &= 6.00\times 10^{-2}\; \rm L\end{aligned}$.

$\begin{aligned}&n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \times V(\mathrm{HCl}) \\ &= 0.750\; \rm mol \cdot L^{-1} \times 6.00\times 10^{-2}\; \rm L \\ &\approx 4.50 \times 10^{-2}\; \rm mol\end{aligned}$.

How many $\rm Zn$ atoms would react that much $\rm HCl$? Notice the ratio between the coefficients of these two species in the balanced chemical equation for this reaction:

$\displaystyle \frac{n(\mathrm{Zn})}{n(\mathrm{HCl})} = \frac{1}{2}$.

In other words, it would take two $\rm HCl$ molecules to react with one $\rm Zn$ atom.

Hence, $4.50 \times 10^{-2}\; \rm mol$ of $\rm HCl$ molecules would react with up to $(1/2) \times 4.50\times 10^{-2}\; \rm mol = 2.25 \times 10^{-2}\; \rm mol$ $\rm Zn$ atoms.

However, only approximately $2.05 \times 10^{-2}\; \rm mol$ of $\rm Zn$ atoms were available in this question. Therefore, $\rm Zn\!$ would be the limiting reactant of this reaction. Up to $2.05 \times 10^{-2}\; \rm mol\!$ of $\!\rm Zn$ atoms would take part in this reaction.

Calculate the per-mole enthalpy change

Assume that this reaction was complete. In other words, assume that all the $\rm Zn$ atoms that could be consumed were actually consumed. Under that assumption, approximately $2.05 \times 10^{-2}\; \rm mol$ of $\rm Zn\!$ atoms had reacted to produced $3.14\; \rm kJ$ of heat through this reaction.

All other things equal, the quantity of heat produced in this reaction should be proportional to the number of $\rm Zn$ atoms that were actually consumed.

Therefore, based on data from this question, the maximum amount of heat that $1\; \rm mol$ of $\rm Zn$ atoms would produce through this reaction would be approximately:

$\begin{aligned}\frac{3.14\; \rm J}{2.05 \times 10^{-2}} \approx 1.53\times 10^{2}\; \rm J\end{aligned}$.