What is the standard enthelpy of formation of NaF

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To determine the standard enthalpy of formation of NaF, first convert the mass of Na (0.560g) to moles using its molar mass (approximately 23 g/mol), resulting in 0.0243 moles of Na. Since the reaction produces NaF in a 1:1 ratio, 0.0243 moles of NaF are formed. The heat evolved is 13.8 kJ, so the standard enthalpy of formation of NaF is calculated by dividing the heat by the moles of NaF produced, yielding approximately -568 kJ/mol. This value aligns with the provided equation for standard enthalpy.
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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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