Heat produced when magnesium is reacted with acid


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> Calculate the heat produced when a strip of $\ce{Mg}$ metal with a mass of $\pu{0.0801 g}$ is reacted with $\pu{50.0 ml}$ of $\pu{1.0 M}$ $\ce{HCl}$ to raise the temperature of the water by $\pu{7.6 °C}$ (change in temperature is $\pu{7.6 °C})$ Calculate the heat produced when a mole of this metal is used.
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I have this so far:


$$q = mc\,\Delta T$$


$m =\pu{0.0801 g}$ and the change in temperature is


$$\Delta T = T\_2 - T\_1 = \pu{7.6 °C} - \pu{4.184 K} = \pu{3.416 K}.$$


Is that right? I have no idea what the specific heat $c$ is. Is it the specific heat of $\ce{Mg}?$


What do I do with the $\pu{50.0 ml}$ of $\pu{1.0 M}$ $\ce{HCl}?$




It can help to imagine the experimental setup.


Imagine a thermally insulated reaction container, let's call it *calorimeter*, that is filled with 50 ml 1 M $\ce{HCl}$ ( = a solution of hydrogen chloride gas in **water**).


We have to measure a temperature change ($\Delta{} T$), that comes from an exothermic reaction. **Where** would we best measure this temperature change? Our **reaction medium** apparently is a good choice. Therefore, there's most likely a thermometer (insulated from the outside) in good thermal contact with the **water**.


Now we let the magnesium strip fall into the **reaction medium** (using some clever mechanic) while our *calorimeter* is closed), let it react and measure the temperature change until the temperature reaches some final (maximum) value.


As you wrote, the temperature change ($ \Delta{} T$) is +7 °C.


I don't see that an initial temperature was given, but that's not important. We have the $ \Delta{} T$.


Now we need a *specific factor* that correlates the heat liberated in the reaction with the temperature change in a *specific medium*.


Here, the *specific heat capacity* of **water** comes into play.


There you go.


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As user user2617804 noted the problem doesn't appear to be well thought out.


The major problem here is that converting mass of magnesium to amount of substance gives


$$n(\ce{Mg}) = \frac{\pu{8.01E-2 g}}{\pu{24.3 g mol^-1}} = \pu{3.30E-3 mol}$$


So, $\pu{50 mL}$ of $1.00$ molar acid only has $\pu{5.0E-2 mol}.$ That little acid can't react with $\pu{1.00 mol}$ of magnesium, so there will be $0.95$ excess moles of magnesium.


Since the magnesium is in excess, Using proportions if $\pu{3.30E-3 mol}$ of acid raises the temperature $7.6$ degrees, then $\pu{5.0E-2 mol}$ of acid should raise the temperature by


$$\Delta T = \frac{\pu{5.0E-2 mol}}{\pu{3.30E-3 mol}}\times \pu{7.6 °C} = \pu{115 °C}$$


So even if the water started at $0\text{ }^\circ\text{C}$ that much heat would boil some of the water off.


The other problem here is that the specific heat of the $\pu{0.95 mol}$ $(\pu{23.1 g})$ of magnesium which would be left is not the same as the specific heat of water.


To really solve this problem, we require


1. initial temperature of water;
2. initial temperature of magnesium metal;
3. specific heat of water;
4. specific heat of magnesium.


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