# Heat produced when magnesium is reacted with acid

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.

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.

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.