Question

How would you calculate the amount of the heat transferred when $3.55{\text{ }}g$ of magnesium,$Mg\left( s \right)$ react at constant pressure? How many grams of magnesium oxide i.e., $MgO$ are produced during an enthalpy change of $- 234{\text{ }}kJ$ ?
Consider the following reaction;
$2Mg\left( s \right) + {\text{ }}{O_2}\left( g \right)\,\, \to {\text{ }}2MgO\left( s \right)\;\;\;\Delta H = - 1204{\text{ }}kJ$

Answer 1

It is a stoichiometry problem which involves the enthalpy as well. The way to solve these problems just by simply treating the enthalpy like another product.
For every $2\,\;mol$ of $Mg$ that react with $1{\text{ }}mol$ of ${O_2}$ ;
We get $2\,\;mol$ of $MgO$ and release $- 1204{\text{ }}kJ$ of heat (since, the energy is released that’s the reason enthalpy is negative).

Complete step-by-step answer: The given a mass of $Mg\left( s \right)$ is $3.55{\text{ }}g$ .
First, we need to convert into moles.
Therefore, the number of moles of $Mg$ will be;
$n\, = \,\dfrac{{mass}}{{molecular\,mass}}$
$= \dfrac{{3.55}}{{24.31\,g/mol}}\,$
$= \,0.146\,mol\,Mg$
Now, we need to find enthalpy.
We know that for every$2\,\;mol$ of $Mg$ , we get an enthalpy release i.e., $- 1204{\text{ }}kJ$ .
So, $Mg$ and Enthalpy are related in a $2:1$ ratio. Keeping this in mind:
$\Delta H\, = \,\dfrac{{0.146}}{2}\, \times \,( - 1204)\,kJ\,$
$\, = \, - 87.91$
So, the enthalpy change is $- 87.91kJ$ , or the amount of heat released is $87.91kJ$ .
In the next step, we do the same thing, but we need to do in backwards.
First, let's talk about the relationship between $MgO$ and enthalpy.
For every $2\,\;mol$ of $MgO$ , you release $- 1204{\text{ }}kJ$ of heat
However, we didn't release $- 1204{\text{ }}kJ$ of heat but we released $- 234{\text{ }}kJ$ of heat.
So, we are going to need to get a ratio, which we can then play with in the reaction:

$\dfrac{{ - 234}}{{ - 1204}}\, = \,0.194$
Now, all we need to do is simply multiply $0.194$ times $2\,$ to find moles of $MgO$ ;
$0.194\, \times \,2\, = \,0.389\,mol\,Mg$
Therefore, $\,0.389\,$ moles of $Mg$ .
We need grams.
So, have to multiply by the molar mass of $MgO$ ;
$(0.389\,mol)(40.31\,g/mol)\, = 15.7\,g\,MgO$

Therefore, the answer is $15.7\,g\,MgO$ .

Note: When a process occurs at constant pressure, the heat gets evolved (either it releases or absorbs) that will be equal to the change in enthalpy. Enthalpy $(H)$ is the sum of the internal energy $(U)$ and the product of pressure and volume $\left( {PV} \right)$ which is given in the form of an equation. Enthalpy is the state function.
$H = U + PV$
Where, $H =$ Enthalpy
$U =$ Internal energy
$P =$ Pressure
$V =$ Volume