Now that you have the amount of fluorine gas in moles, you can start the stoichiometry problem.

0.895 mol F2 × (2 mol NaCL/1 mol F2) × (58.44 g NaCl/1 mol NaCl) = 105 g NaCl

Why didn't they convert the Flourine?

Question

Now that you have the amount of fluorine gas in moles, you can start the stoichiometry problem.

0.895 mol F2 × (2 mol NaCL/1 mol F2) × (58.44 g NaCl/1 mol NaCl) = 105 g NaCl

Why didn't they convert the Flourine?

xishem · Answer

We need more context. Could you post more of the problem? At least the original problem would help.

xishem · Answer

What exactly are you confused about? I'm going to assume that this is the reaction we are looking at...

$$F_2+ 2 NaCl = 2NaF + Cl_2$$In this case, I'm going to guess that the question is asking what mass of NaCl will be consumed when it reacts with 0.895mol of fluorine gas (assuming excess NaCl).

You use a mole conversion factor: for each mole of fluorine gas that is consumed, 2 moles of sodium chloride will be consumed. Then that amount of sodium chloride is simply converted to a mass of NaCl.