Answer:
Moles of NOâ‚‚ = 0.158
Explanation:
             SO 2 ( g ) + NO 2 ( g ) ⇄  SO 3 ( g ) + NO ( g )
       According to the law of mass equation
                  Â
                    [tex]K_{c}[/tex] = [tex]\frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2} ]}[/tex]
               ⇒  3.10 = [tex]\frac{(1.00)(1.00)}{(2.30) [NO_{2} ]}[/tex]   At equilibrium [SO₃] = [NO]
               ⇒ [NO₂] = [tex]\frac{1}{6.3}[/tex]
               ⇒ [NO₂] = 0.158
So. number of moles of NOâ‚‚ at equilibrium added = 0.158