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When 0.45 g of Zn is added to 50.0 ml of 0.95 M HCl solution, the solution inside the calorimeter heats up by 12 °C. What is the molar enthalpy of the reaction (in kJ/mol)?

I know how to find the moles using $n=cV$ and $n=m/M$ but I'm not sure I understand why when doing $mc\Delta T$ we use the specific heat capacity of water when the solution is HCl.

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  • $\begingroup$ Welcome to Chem SE! If you haven't yet, take the tour and visit the help center. A solution of HCl is not liquid HCl, but rather HCl dissolved in water, so for a dilute enough solution you can use the heat capacity of pure water. $\endgroup$ – Tyberius Jan 26 at 19:01
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Amount of $\ce{Zn}$: $$n_{\ce{Zn}}=\frac{0.45\ \mathrm g}{65\ \mathrm{ g/mol}}=0.0069\ \mathrm{mol}$$

Amount of $\ce{HCl}$: $$n_{\ce{Zn}}=MV=0.95\ \mathrm{mol\ l^{-1}}\times\frac{50\ \mathrm{ml}}{1000\ \mathrm{ml/l}}=0.0475\ \mathrm{mol}$$ The reaction $$\ce{Zn + 2HCl -> ZnCl2 + H2}$$

So $0.0069\ \mathrm{mol}$ zinc requires $2\times0.0069\ \mathrm{mol}=0.0138\ \mathrm{mol}$ HCl; thus $\ce{Zn}$ is the limiting reagent here which decides the extent of the reaction.

Approximating specific heat capacity of solution to be specific heat capacity of water: $c_p=4.2\ \mathrm{J\ ^\circ C^{-1}\ g^{-1}}$.

Net heat released: $$Q=mc_p(T_2-T_1)=50\ \mathrm g\times4.2\ \mathrm{J\ ^\circ C^{-1}\ g^{-1}}\times12\ \mathrm{^\circ C}=2520\ \mathrm J$$

So heat release per amount of zinc: $$Q=\frac{2520\ \mathrm J}{0.0069\ \mathrm{mol}}=365217\ \mathrm{J/mol}=365\ \mathrm{kJ/mol}$$

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