Question

How many liters of fluorine gas can react with 32.0 grams of sodium metal at standard temperature and pressure? Show all of the work used to find your answer.

2Na + F2 yields 2NaF

Answer 1

I have:
32 g NaF x \[\frac{ 1 mol NAF }{ 41.98 g NaF}\] x\[\frac{ 1 mol F _{2} }{ 2 mol NaF }\] x \[\frac{ 1 L F _{2} }{ 1 mol F _{2} }\] = .381

Answer 2

Is that correct?

Answer 3

@gabylovesu

Answer 4

@karatechopper

Answer 5

@mathstudent55

Answer 6

@FutureVet

one small mistake

the last step should be 22.4L F2/1 mol F2, because 1 mol of gas at standard temperature and pressure occupies 22.4L

Answer 7

other than that, everything else looks ok

Answer 8

Thanks I realized that before I Submitted it.