Adding $\pu{2.00g}$ $\ce{Mg}$ metal to $\pu{95.0 mL}$ of $\mathrm{1.00 M}$ $\ce{HCl}$ in a coffee-cup calorimeter leads to an increase of $\pu{9.2 ^\circ C}$. If the molar heat capacity of $\pu{1.00 M }$ $\ce{HCl}$ is the same as that for water $(\pu{C_p = 75.3 J mol-1 K-1)}$, what is the heat of reaction?

Two relevant equations are: $${\Delta H_{(rxn)} = -q}\\ {q = n \times C_p \times \Delta \mathrm T}$$

The answer given is $\pu{-44.4 kJ mol-1}$. I wonder if I'm not way off the mark here by using the wrong equations, because the answer I got was $\pu{122.83 J mol-1 \; ( 0.1773 mol \times 75.3 J \times 9.2)}$.

note: The heat of reaction = $\Delta H_{(rxn)}$


1 Answer 1


The equations you listed can be used to partially solve the problem. However, with the equations you used, you have made one error which was in your choice of "moles." Your moles value must match up with your $\ce{C_{p}}$, so you must find the moles of Water in $\pu{95.00mL}$! Alternatively, you can use $4.184 \pu{\frac{J}{g\ ^\circ C}}$ and $\pu{1 \frac{g}{mL}}$ and $\pu{95.00mL}$ of water along with the equation $\mathrm{Q = n \times C_p \times \Delta T}$.

On the other hand, the premise of your solution is close but wrong, which is obvious if you examine the units:

$\pu{0.1773 mol} \times 75.3 \frac{\pu J}{\pu{mol K}} \times \pu{9.2 K} = \pu{122.8 J}$

The units for Heat of Reaction are $\pu{\frac{J}{mol}}$, not $\pu{J}$!

To finish off your solution, you must divide by the moles of reaction that occurred (given by the limiting reactant). $\pu{-44.4 kJ}$ is the value calculated based on the moles of Mg reacted. However, I suspect that the given answer is wrong because the balanced equation does not yield a 1:1 reaction ratio between $\ce{Mg}$ and $\ce{HCl}$, which means that $\ce{HCl}$ is the limiting reactant, which would make the Heat of Reaction Larger.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.