The standard enthalpy of formation,
In your case, the most important thing to remember here is that you're looking for the enthalpy change of reaction that corresponds to the formation of one mole of the compound.
Sodium metal will react with chlorine gas to form sodium chloride according to the balanced chemical equation
#"Na"_text((s]) + 1/2"Cl"_text(2(g]) -> "NaCl"_text((s])#
So, you have a
Since chlorine gas is said to be in excess, you can assume that all the sodium metal will actually react. Use sodium metal's molar mass to determine how many moles you have in that
#11.5color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23.0color(red)(cancel(color(black)("g")))) = "0.50 moles Na"#
Now, you know that when
This means that when
#1color(red)(cancel(color(black)("mole NaCl"))) * "205.5 kJ"/(0.50color(red)(cancel(color(black)("moles NaCl")))) = "411 kJ"#
It's very important to remember that when heat is being given off, the enthalpy of reaction carries a negative sign.
This negative sign is used to symbolize the fact that the system is losing heat. This means that the standard enthalpy change of formation for sodium chloride will be
#DeltaH_f^@ = color(green)(-"411 kJ mol"^(-1))#
The answer is rounded to three sig figs, the number of sig figs you have for the mass of sodium.
This value tells you that when one mole of sodium chloride is formed from its constituent elements in their stable form,