NaNO2 and 11.5g are obtained from heating 75.1g NaNO3, what is the mass percent yield of O2?

In the question, "…11.5 g are obtained…" should be "…11.5 g of O₂ are obtained" instead.

Molar mass of NaNO₃ = 23.0 + 14.0 + 16.0×3 = 85.0 g/mol
Molar mass of O₂ = 16.0×2 = 32.0 g/mol

2NaNO₃ → 2NaNO₂ + O₂
Molar ratio NaNO₃ : O₂ = 2 : 1

Moles of NaNO₃ = (75.1 g) / (85.0 g/mol) = 0.8835 mol
Maximum moles of O₂ obtained = (0.8835 mol) × (1/2) = 0.4418 mol
Theoretical yield of O₂ = (0.4418 mol) × (32.0 g/mol) = 14.1 g

Percent yield of O₂ = (11.5/14.1) × 100% = 81.6%

====
OR:
(75.1 g NaNO₃) × (1 mol NaNO₃ / 85.0 g NaNO₃) × (1 mol O₂ / 2 mol NaNO₃) × (32.0 g O₂ / 1 mol O₂)
= 14.1 g O₂ (theoretical yield)

(11.5/14.1) × 100% = 81.6% yield

75.1 X 69/83  =  62.43  <<<  theoretical yield

   11.5/62.43  X 100 = 18.4%