Consider the following equilibrium: 2NOCl( g ) lt;—–gt; 2NO( g ) + Cl 2 ( g ) K = 1.6 X 10 -5 1.00 mole of pure NOCl and 0.958 mole of pure Cl 2…

2NOCl(g)  <—–> 2NO(g) + Cl2(g)         K = 1.6 X 10–5

1.00 mole of pure NOCl and 0.958 mole of pure Cl2 are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g).

Provide all work and equations used for correct answer.