Question

Standard heat of formation of HgO(s) at 298 K and at constant pressure 8s -90.8 kJ/mol. If excess of HgO(s) absorbs 41.84 kJ of heat at constant volume, calculate the amount of Hg that can be obtained at constant volume and 298 K. (At wt. of Hg=200)​

Answer 1

Answer:

The amount of Hg that can be obtained at constant volume and 298 K is 0.4608 g.

Explanation:

$Hg(l)+\frac{1}{2}O_2(g)\rightarrow HgO(s),\Delta H_f=-90.8 kJ/mol$

Standard heat of formation of HgO(s) = $\Delta H_f=-90.8 kJ/mol$

Amount of heat absorbed by HgO , Q= 41.84 kJ

$HgO(s)\rightarrow Hg(l)+\frac{1}{2}O_2(g),\Delta H=90.8 kJ/mol$

Moles of HgO= $\frac{Q}{\Delta H}=\frac{41.84 kJ}{90.8 kJ/mol}=0.4608 mol$

According to reaction 1 mol of HgO gives 1 mol of Hg.

then 0.4608 mol of HgO will giove:

$\frac{1}{1}\times 0.4608 =0.4608 mol$ of Hg.

Mass of 0.4608 mol of Hg = 0.4608 mol × 200 g/mol = 92.16 g