## anonymous one year ago 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

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1. anonymous

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

2. anonymous

Is that correct?

3. anonymous

@gabylovesu

4. anonymous

@karatechopper

5. anonymous

@mathstudent55

6. Vocaloid

@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

7. Vocaloid

other than that, everything else looks ok

8. anonymous

Thanks I realized that before I Submitted it.