What mass (in g) of oxygen must be added to 8.89 g of neon at 23.6oC, and in a 103 L contain for a final pressure of 850 mmHg?

I did PV=nRT and converted with molar mass and got 151.4 but its not the correct answer

You had a calculation error

V = 103 L
T = (273.15 + 23.6)K
R.gas = 8.314 J / mol*K

8.89 gm Neon = .441mol

Pressure of Neon = (.441)*R.gas*T/V

850 mmHg - Pressure of Neon = Pressure of Oxygen

mols of Oxygen = (Pressure of Oxygen) * V/ (R.gas * T)

Mass of Oxygen = Mols of Oxygen * 32gm

I got 137.296gm

Thank you!