Help me please?

Question

When 0.560 g 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 (H°f) of NaF(s)?

Na(s) + F2(g)  NaF(s) (H°f) ??

A. 24.8 kJ/mol B. 570 kJ/mol C. 24.8 kJ/mol D. -570 kJ/mol

Answer 1

You know how much heat is evolved using 0.560 g. What you need is how much heat per mole NaF. But one mole NaF is formed from one mole Na.

So this is just a scaling problem. Convert 0.560 g Na to moles Na, divide heat evolved in expt by moles Na in expt,and there you have heat evolved per mole, which is what you want.

You want a negative number as your answer, of course (why?)