# 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?

Jul 2, 2016

$\text{30.3 kJ}$

#### Explanation:

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

$\frac{1}{2} \text{O"_ (2(g)) + "Hg"_ ((l)) -> "HgO"_ ((s))" "DeltaH = -"90.8 kJ}$

This equation tells you that when $1$ mole of mercury reacts with $\frac{1}{2}$ moles of oxygen gas, $\text{90.8 kJ}$ 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, $\Delta H$, symbolizes heat given off.

So, you now want to know how much heat will be released when $\text{66.9 g}$ 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 $\Delta H$ 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 $\text{66.9 g}$ of mercury undergo combustion, the enthalpy change of reaction is

$\Delta {H}_{\text{rxn" = -"30.3 kJ}}$

Once again, the minus sign symbolizes heat given off.