# Heat Produced when Mg is reacted with HCL

Calculate the heat produced when a strip of $\ce{Mg}$ metal with a mass of $0.0801$ gram is reacted with $50.0$ml of $1.0M$ $\ce{HCl}$ to raise the temperature of water to $7.6^\circ C$ (change in temp.=$7.6^\circ C$). Calculate the heat produced when a mole of this metal is used.

I have this so far $q= m \times c \times \Delta T$

$m=0.0801$gram
change in temp($\Delta T$)= ($7.6-4.184=3.416$) ($T_f-T_i$) is that right? and I have no idea what the specific heat($C$) is.

is it the specific heat of Mg? what do I do with the $50.0$ml of $1.0M$ $\ce{HCl}$? thanks

• Good start Vivian. The chemistry problem statement is no well thought out but I believe C is the heat capacity of water (close to reality). You should start with the chemical reaction between Magnesium and hydrochloric acid which produces a magnesium chloride solution and hydrogen. – user2617804 Jan 31 '14 at 5:00

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 grams of magnesium to moles gives

moles Mg = $\dfrac{0.0801 \text{ g}}{24.3 \text{ g/mole}} = 0.00330\text{ moles}$

So 50 mL of 1.00 molar acid only has 0.050 moles. That little acid can't react with 1.00 mole of magnesium, so there will be 0.95 excess moles of magnesium.

Since the magnesium is in excess, Using proportions if 0.00330 moles of acid raises the temperature 7.6 degrees, then 0.050 moles of acid should raise the temperature

$\Delta T=\dfrac{0.050}{0.00330}\times7.6 \text{ }^\circ\text{C} = 115 \text{ }^\circ\text{C}$

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 0.95 moles (23.1 grams) of magnesium which would be left is not the same as the specific heat of water.

So to really solve this problem requires (1) initial temperature of water, (2) initial temperature of magnesium metal (3) Specific heat of water, and (4) specific heat of magnesium.