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

32 g Na x (1 mol Na / 23.0 g Na) x (1 mol F2 / 2 mol Na) x (22.4 L / 1 mol F2) = 15.6 L of fluorine 1.5 x 2.0 = 12.0 x Volume = 0.25 L

Answer 2

15.58 L of F₂

Solution:

The Balance Chemical equation is as follow,

2 Na + F₂ → 2 NaF

As we know at Standard temperature and pressure 1 mole of ideal gases occupies 22.4 L of volume. Let us assume that in our case the Fluorine gas is acting ideally.

So, According to equation,

46 g (2mol) Na reacts with = 22.4 L (1 mol) F₂

So,

32 g Na will react with = X L of F₂

Solving for X,

X = (32 g × 22.4 L) ÷ 46 g

X = 15.58 L of F₂