2NO(g) + Br2(g) ⇌ 2NOBr(g)
According to this equation, 2 moles of NO(g) react with 1 mole of Br2(g) to form 2 moles of NOBr(g). The composition of the equilibrium mixture can be calculated as follows:
No. of moles of NOBr(g) formed at equilibrium = 0.0518 mol
No. of moles of NO(g) taking part in reaction = 0.0518 mol
No. of moles of NO(g) left at equilibrium = 0.087 - 0.0518 = 0.0352 mol
No. of moles of Br2(g) taking part in reaction = \(\frac{1}{2}\) x 0.0518 = 0.0259 mol
No. of moles of Br2(g) left at equilibrium = 0.0437 - 0.0259 = 0.0178 mol
Hence 0.0352 mol of NO and 0.0178 mol of Br2