srry I placed the question in the wrong spot.
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.
To find the concentration of NH3 at equilibrium, we need to use the stoichiometry of the balanced equation.
From the given equation:
2NH3 <=> N2 + 3H2
We see that 2 moles of NH3 are required to produce 1 mole of N2.
Given that the concentration of N2 at equilibrium is 0.1 M, we can use this information to calculate the amount of NH3 used.
0.1 M N2 x (2 mol NH3 / 1 mol N2) = 0.2 mol NH3 used
Alternatively, we are also given that 0.3 moles of H2 were produced. From the balanced equation, we can see that 3 moles of H2 are produced from 2 moles of NH3.
So, we can calculate the amount of NH3 used using the amount of H2 produced:
0.3 mol H2 x (2 mol NH3 / 3 mol H2) = 0.2 mol NH3 used
Either way, we find that 0.2 moles of NH3 were used.
To find the concentration of NH3 at equilibrium, we started with 1 mole of NH3 and used 0.2 moles. Therefore, the remaining amount of NH3 at equilibrium is 1.0 - 0.2 = 0.8 moles.
Since the total volume of the flask is 1 liter, the concentration of NH3 at equilibrium is:
Concentration of NH3 = Amount of NH3 / Volume of flask
= 0.8 mol / 1 L
= 0.8 M
Therefore, the concentration of NH3 at equilibrium is 0.8 M.