I've been looking at this and it is a bit fiddly - there is probably an easier way to do it, but here's what I did:
As Borek said, since we have 0.1 mol of potassium, we must have 3.91 g of potassium.
That leaves us with 7.63-3.91 = 3.72 g of halide.
We know that one of the halides must be F
- because, as Borek also pointed out, the mass of the precipitate is too low to contain 0.1 mol of silver, so one of the silver halides must be soluble - AgF is the only soluble silver halide.
So, where m
A = mass of element A, X is the unknown halogen and M
r is relative mass of unknown halogen:
m
F + m
X = 3.72 (equation 1)
We also know that if we have 0.1 mol of potassium, we must have 0.1 mol of halides:
m
F/19 + m
X/M
r = 0.1 (equation 2)
Considering the precipitate 5.64 g of AgX:
m
X + m
Ag = 5.64 (equation 3)
Since the moles of Ag = moles of X:
m
Ag/108 = m
X/M
ror m
Ag = 108m
X/M
r (equation 4)
Substitute equation 4 into equation 3 and we get:
m
X + (108m
X/M
r) = 5.64
Which rearranges to:
M
r = 108m
X/(5.64-m
X) (equation 5)
Substitute equation 5 into equation 2 and we get:
m
F/19 + [(5.64 - m
X)/108] = 0.1
Which rearranges to:
5.68m
F - m
X = 5.16 (equation 6)
Equations 6 and 1 can be solved (
simultaneous equations) to find the values of m
F and m
X.
Once you know these values you can use equation 2 to determine M
r - compare this value to your periodic table to identify the halogen.
Like I say, this is really long winded and there is probably an easier more obvious approach to this problem that I've missed.