I need help with equilibrium in ap chemistry please help?

3H2 + N2 > 2NH3

Initial conc. - (H2=4)  (N2=3)  (NH3=0)
Change Rxn- (-3x) (-x) (+2x)
Equilibrium - (0.7)

H2 -- 4-3x=.7
x=1.1

[N2] = 3-1.1=1.9 M
[NH3] = 0+2(1.1) = 2.2 M

Kc = ([NH3]^2) / ([H2]^3[N2]) = (2.2^2) / (.7^3 x 1.9) = 4.84/(.6517) = 7.43

Lucky for you, we *just* finished the equilibrium unit

Set up an ICE table
The first column is [H2], the second is [N2] and the third is [NH3]

Initial Concentration------------------ 4 M-------------3M------------0
Change in Concentration~~~~~~-3x~~~~~~~~-x~~~~~+2x
Equilibrium Concentration~~~~~4-3x~~~~~~~3-x~~~~~2x

You know what x is based on the number of moles of substance being used/formed

4-3x = 0.7
x = 1.1

[H2] = 0.7M
[N2] = 1.9M
[NH3] = 2.2M

Kc = products/reactants
Kc = [NH3]^2 / ([H2]^3 x [N2])
= (2.2)^2 / (0.7)^3 x 1.9
= 7.43

Hopefully, I haven't made some silly arithmetic error and what I've done is clear enough for you.