Answer: 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.
