srry I placed the question in the wrong spot.

Question

The equation is
2NH3 <=> N2 + 3H2
1 mole of NH3 injected intoa 1L flask
0.3 moles of H2 was found
the concentration of N2 at equlibrium is 0.1M

how do I find the concentration of NH3 at equlibrium?

(N2) = 0.1
(H2) = 0.3
Amount NH3 used is 0.1 mol N2 x (2 mol NH3/1 mol N2) = 0.1 x 2/1 = 0.2 mol NH3 used.

OR you can start with 0.3 H2 like this;
0.3 H2 x (2 mol NH3/3 mol H2) = 0.2 mol NH3 used.

We started with 1 mol NH3; therefore, we have remaining 1.0-0.2 = 0.8 mol/L.