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

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.

Best regards.

Answer:

83.5 kJ

Explanation:

Molar mass is 146.053. Moles of 10.0 g = 0.0685 mol. Multiply by 1220 kJ/mol for the answer of 83.5 kJ.

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