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Correct Answer: B) 0.75 g NO and 0.45 g O₂
560 cm³ = 0.560 L. Moles of NO₂ = 0.560/22.4 = 0.025 mol. Thus NO consumed = 0.025 mol = 0.75 g. O₂ consumed = 0.0125 mol = 0.40 g. Unused O₂ = 50 mg = 0.05 g. Initial O₂ = 0.40 + 0.05 = 0.45 g.