Answer:
Moles of NO₂ = 0.158
Explanation:
SO 2 ( g ) + NO 2 ( g ) ⇄ SO 3 ( g ) + NO ( g )
According to the law of mass equation
[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 NO₂ at equilibrium added = 0.158