How many moles of NaF are produced in the resction between sodium bromide and calcium fluoride when 550 grams of sodium bromide are used

5 moles of NaF would be produced when 550 grams of Sodium bromide are used.

Explanation:

From the question, the balanced chemical formula is illustrated below:

2NaBr + CaF2 → 2NaF + CaBr2

From the equation above;

Molecular mass of NaBr = 23 + 80 = 103g/ mol

Molecular mass of CaF2= 40+ (19×2)= 78g/ mol

Molecular mass of NaF = 23+19 = 42g/ mol

From the balanced chemical equation;

2 moles of NaBr reacted with 1mole of CaF to give 2 moles of NaF.

That is, 2 moles of NaBr= 2× 103 = 206g

2moles of NaF = 2×42= 84g

If 206g of NaBr yielded 84g of NaF

Therefore 550g of NaBr will yield Xg of NaF

X= 550×84/206

X= 224.27g of NaF

But 42g = 1 mole of NaF

Therefore 224.27g = X mole of NaF

X= 224.27 ×1/42

X is approximately 5moles.

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.