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 answer is barium and copper bromide