How do you calculate the equilibrium amounts of H2 and HF?

You can set up something that can be called an ICE table. I stands for initial, C stands for change, and E stands for equilibrium. All the amounts are calculated in Molarity.
So you have the equation: H2(g) + F2(g) –> 2HF(g)
So you can set up the table in the following way:

Molarity:____[H2]_____[F2]_____[HF]
I _________ .1 _____ .1________0
C__________- X______- X______+ 2X
E_________.1 – X____.1 – X______2X

For H2 and F2 you know that it’s minus X because the initial amount is decreasing. Also, the coefficient on the X is the same as the coefficient in the balanced chemical equation.

Now, the equation for the equilibrium constant, K, for the generic chemical equation aA + bB –> cC + dD is:

K = ([C]^c[D]^d) / ([A]^a[B]^b)
where the square brackets mean Molarity. So now substitute in your values.

1.0 x 10^2 = [2X]^2 / [.1 - X] [.1 - X]
Now you could solve this in two ways. One is to use the quadratic formula to solve for X, and then substitute the value in to the ICE table to get your values, or you could assume X is small, and can be discounted. This means you can throw out X in any term where it is being added or subtracted, but no other time. Solving also, X can never be negative, so if you’re solving the quadratic formula and you get a negative value, just throw it out and use the other one.