the problem states:
a compound of chlorine and fluorine, ClFn, reacts at about 75 degrees C with uranium metal to produce uranium hexafluoride, UF6, and chlorine monofluoride, CIF(g). a quantity of uranium produced 3.53 g UF6 and 343 mL ClF at 75 degrees C and 2.50 atm. what is the formula (n) of the compound?

i got the moles of ClF(g) = 0.30 mol
and the moles of UF6 = 0.01mol

but how do i get the formula of ClFn after that? i know I'm supposed to get the mol of F from the two products, but how?? i really want to understand this

the answers supposed to be ClF3.

Moles of  F in ClF =0.3
Moles of  Cl in ClF =0.3
Moles of  F in UF6 =0.01X6 = 0.06
total number of moles of F = 0.06 + 0.3 =0.36
so in the given compound Cl anf F are present as 0.3:0.36= 1: 1.2
this is how the calculations should be done
I dont think answer provided by you is correct