Answer
See the answer below.
Work Step by Step
a)
$\begin{smallmatrix}
CaBr_2(s)+H_2O(g)&\rightarrow &CaO(s)+2\ HBr(g)\\
HgBr_2(s)+\underline{CaO}(s)&\rightarrow &HgO(s)+\underline{CaBr_2}(s)\\
\underline{HgO}(s)&\rightarrow&{Hg}(l)+1/2\ O_2(g)\\
\underline{Hg}(l)+\underline{2\ HBr}(s)&\rightarrow &\underline{HgBr_2}(s)+H_2(g)\\
\\
\hline\\
H_2O(g)&\rightarrow&H_2(g)+1/2\ O_2(g)
\end{smallmatrix}$
b) From the overall stoichiometry:
$1000\ kg\div 18.015\ kg/kmol\ H_2O\times 1\ kmol\ H_2/kmol\ H_2O\times2.016\ kg/kmol\ H_2=112\ kg$
c) Enthalpies of formation (kJ/mol):
$CaO(s): -635.09, H_2O(g): -241.83, HBr(g): -36.29, HgO(s): -90.83$
First reaction:
$-635.09+2\times -36.29--241.83--683.2=+217.36\ kJ/mol$
Endothermic
Second reaction:
$-169.5-2\times-36.29=-96.92\ kJ/mol$
Exothermic
Third reaction:
$-683.2-90.83--635.09--169.5=+30.56\ kJ/mol$
Endothermic
Fourth reaction:
$--90.83=+90.83\ kJ/mol$
Endothermic.
d) This method is subject to a series of industrial inefficiencies and logistical problems, which could be worth it in a cost-benefit analysis, but an alternative method, like electrolysis, should be considered.
