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

2NaBr+CaF2 -> 2NaF+CaBr2

A.0.53 mol NaF
B. 5g NaF
C. 10.7 mol NaF
D. 5.35 g NaF

Respuesta :

sebassandin

Answer:

D. 5.35 g NaF.

Explanation:

Hello!

In this case, given the chemical reaction, whereas there is a 2:2 mole ratio between NaF and NaBr, we can compute the moles of NaF product by firstly computing the moles of NaBr in 550 and secondly using the mole ratio as shown below:

[tex]n_{NaF}=550gNaBr*\frac{1molNaBr}{102.89gNaBr}*\frac{2molNaF}{2molNaBr} \\\\n_{NaF}=5.35molNaF[/tex]

Therefore, the answer is D. 5.35 g NaF.

Best regards!

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