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

Hint: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