Question

What heat is liberated in the formation of 10.0 grams of sulfur hexafluoride, SF6, from the elements sulfur and fluorine?

Answer 1

Answer:

$Q=-76.7kJ$

Explanation:

Hello,

In this case, for such formation of sulfur hexafluoride, the standard enthalpy of formation is -1220.47 kJ/mol (data extracted from NIST database). Next, we compute the moles in 10.0 grams of sulfur hexafluoride as shown below:

$n_{SF_6}=10.0gSF_6*\frac{1molSF_6}{146.06 gSF_6}=0.0685mol$

Next, for the given energy, we compute the total heat that is liberated:

$Q=-1220.47\frac{kJ}{mol}*0.0685 mol\\\\Q=-76.7kJ$

Finally, we conclude such symbol has sense since negative heat is related with liberated heat.

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