When $\pu{20.00 mL}$ of $\pu{1.00 M}$ $\ce{AgNO3}$ solution is added to $\pu{20.00 mL}$ of $\pu{1.00 M}$ of $\ce{NaI}$ at $\pu{25 ^\circ C}$ in a calorimeter, a white precipitate of $\ce{AgI}$ is formed. The temperature of the aqueous mixture increases to $\pu{40 ^\circ C}$.

I am trying to calculate the $\Delta H$ for the reaction per mole of $\ce{AgI}$. The specific heat of the aqueous mixture is $\pu{4.184 J//g K}$, the density of the mixture $\rho = \pu{1.00 g//ml}$. And I assume that the calorimeter absorbs a negligible amount of heat.

How am I suppose to go about finding the $\Delta H$ per mole of $\ce{AgI}$? I don't really know where to start here as I am unclear about the whole problem here. In order to use the formula $$Q = m C_p \Delta T$$ I need to find the mass. But which mass do I take? $\ce{AgI}$, or $\ce{NaNO3}$, or is it the mass of the entire product of the reaction?

This is the balanced equation $$\ce{AgNO3 + NaI -> AgI + NaNO3}$$

  • $\begingroup$ I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. $\endgroup$ Dec 12, 2017 at 16:46

1 Answer 1


When the reaction takes place and releases its heat, everything in the calorimeter heats up: the water, the $\ce{AgI}$ and the $\ce{NaNO3}$. But of these the water is by far the most massive and for that reason, as a reasonable approximation, we consider that mass that absorbs the heat to be the water, i.e. $40.00\ \mathrm{g}$.

Use this to calculate $Q$.

Then determine how many moles of $\ce{AgI}$ was formed, by stoichiometry. The quotient of these two numbers gives you the reaction enthalpy per mole of $\ce{AgI}$

  • $\begingroup$ So, for Q I got 2510.4J (using mass as 40g as we assume no changes in volume. By mole ratio, I got 0.02 Moles AgI . So the question wants per mole of AgI . So am I right to say that 0.02 moles of AgI makes 2510.4 J ? And now I just need to find 1 mole of AgI make much heat in joules ? $\endgroup$
    – user175089
    Dec 13, 2017 at 1:44
  • $\begingroup$ Yes. Note that "we assume no changes in volume" is irrelevant: this is about mass, not volume. $\endgroup$
    – Gert
    Dec 13, 2017 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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