1. Write the reaction

For combustion of mercury vapor to mercuric oxide:

$2\,\mathrm{Hg}(g) + \mathrm{O_2}(g) \rightarrow 2\,\mathrm{HgO}(s)$

We are told that $2$ mol of $\mathrm{Hg}(g)$ reacts.

2. Heat released in bomb calorimeter

At constant volume,

$q_v = \Delta U$

The calorimeter absorbs:

$q_{\text{cal}} = C_{\text{cal}}\Delta T$

Given:

$C_{\text{cal}} = 20.00\,\mathrm{kJ\,K^{-1}}$ $\Delta T = 312.8 - 298.0 = 14.8\,\mathrm{K}$

So,

$q_{\text{cal}} = 20.00 \times 14.8 = 296.0\,\mathrm{kJ}$

Since the reaction is exothermic, the reaction releases this heat:

$q_{\text{rxn}} = -296.0\,\mathrm{kJ}$

Thus for the reaction as written for $2$ mol Hg,

$\Delta U = -296.0\,\mathrm{kJ}$

3. Convert $\Delta U$ to $\Delta H$

Use:

$\Delta H = \Delta U + \Delta n_g RT$

Here,

• Reactant gases: $2\,\mathrm{Hg}(g) + 1\,\mathrm{O_2}(g)$ gives $3$ mol gas
• Product gases: none

So,

$\Delta n_g = 0 - 3 = -3$

Now,

$\Delta H = -296.0 + (-3)(8.3\times 10^{-3})(298)$

because $R=8.3\,\mathrm{J\,mol^{-1}K^{-1}}=8.3\times 10^{-3}\,\mathrm{kJ\,mol^{-1}K^{-1}}$.

Compute:

$(-3)(8.3\times 10^{-3})(298) = -7.42\,\mathrm{kJ}$

Hence,

$\Delta H = -296.0 - 7.42 = -303.42\,\mathrm{kJ}$

This is for:

$2\,\mathrm{Hg}(g) + \mathrm{O_2}(g) \rightarrow 2\,\mathrm{HgO}(s)$

4. Use enthalpy of formation relation

For the reaction,

$\Delta H_{\text{rxn}} = \sum n\Delta H_f^\circ(\text{products}) - \sum n\Delta H_f^\circ(\text{reactants})$

Given:

$\Delta H_f^\circ\big(\mathrm{Hg}(g)\big)=61.32\,\mathrm{kJ\,mol^{-1}}$ $\Delta H_f^\circ\big(\mathrm{O_2}(g)\big)=0$

Let

$\Delta H_f^\circ\big(\mathrm{HgO}(s)\big)=X$

Then for 2 mol HgO formed:

$-303.42 = 2X - 2(61.32)$

$-303.42 = 2X - 122.64$

$2X = -303.42 + 122.64 = -180.78$

$X = -90.39\,\mathrm{kJ\,mol^{-1}}$

Therefore,

$|X| = 90.39$

As an integer-type answer, this is:

$\boxed{90}$

5. Compare with stored correct answer

Stored correct answer range: $89.00$ to $91.00$

Our answer $90.39$ lies in this range, so it agrees.