The #DeltaH# value for the reaction #1/2O_2 (g) + Hg(l) -> HgO (s)# is -90.8 kJ. How many kJ are released when 66.9 g #Hg# is reacted with oxygen?

Start by taking a look at the thermochemical equation given to you

#1/2"O"_ (2(g)) + "Hg"_ ((l)) -> "HgO"_ ((s))" "DeltaH = -"90.8 kJ"#

This equation tells you that when #1# mole of mercury reacts with #1/2# moles of oxygen gas, #"90.8 kJ"# of heat are being given off by the reaction.

Keep in mind that the minus sign used in the expression of the enthalpy change of reaction, #DeltaH#, symbolizes heat given off.

So, you now want to know how much heat will be released when #"66.9 g"# of mercury take part in the reaction. Since you know how much energy is released when #1# mole of mercury reacts, use the element's molar mass to convert the mass to moles

#66.9 color(red)(cancel(color(black)("g"))) * "1 mole Hg"/(200.6color(red)(cancel(color(black)("g")))) = "0.3335 moles Hg"#

You can now use the known #DeltaH# to determine how much heat is released when #0.3335# moles of mercury react

#0.3335 color(red)(cancel(color(black)("moles Hg"))) * "90.8 kJ"/(1color(red)(cancel(color(black)("mole Hg")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("30.3 kJ")color(white)(a/a)|)))#

The answer is rounded to three sig figs.

This is equivalent to saying that when #"66.9 g"# of mercury undergo combustion, the enthalpy change of reaction is

#DeltaH_"rxn" = -"30.3 kJ"#

Once again, the minus sign symbolizes heat given off.