What is the standard enthelpy of formation of NaF

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SUMMARY

The standard enthalpy of formation of sodium fluoride (NaF) is calculated based on the reaction of 0.560g of sodium (Na) with excess fluorine gas (F2), which releases 13.8 kJ of heat. The enthalpy change per mole of NaF produced is given as -569 kJ/mol. To find the standard enthalpy of formation, the number of moles of Na involved in the reaction must be determined first, which is essential for scaling the heat evolved to a per mole basis.

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Homework Statement


Please help! i just need to see one step by step solution of how to solve this problem and then I can figure the rest out.
When .560g of Na (s) reacts with excess F2(g) to form NaF(s), 13.8 kj of heat is evolved at standard state conditions. What is the standard enthalpy of formation of NaF (s)?




Homework Equations


(delta H=-569Kj/Mol)


The Attempt at a Solution

 
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Standard enthalpy is per mole NaF produced, you have it per 0.560g of Na. How many moles produced?
 

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