Answer:

Explanation:
In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.
Here's a balanced equation for the reaction:

Let us define
work as;

where 
External pressure is given as 
, therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence, 
To find the volume of the products, we need to first find the amount of moles of the product made from
using the molar mass of
which is 101.1032 g/mol

Now let us convert moles of
into moles of
and
using the stoichiometric ratios from our balanced equation of the reaction.


is not factored into the volume calculation because it is a solid.
Now let us also convert the moles of
and
into grams using their respective molar masses.


We will now proceed to convert grams into volume using the density values provided.


Summing up the two volumes, we get the final volume

Plugging everything into the
equation, we get:

Finally, let us convert
into joules using the conversion rate of;
