$$4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$$
Next, we calculate the number of moles of ammonia using its molar mass:
$$Moles\ of\ NH_3 = \frac{23.9\ g}{17.04\ g/mol} = 1.404\ mol$$
According to the balanced chemical equation, 4 moles of ammonia react with 5 moles of oxygen. Therefore, we can calculate the moles of oxygen required:
$$Moles\ of\ O_2 = 1.404\ mol\ NH_3 \times \frac{5\ mol\ O_2}{4\ mol\ NH_3} = 1.755\ mol\ O_2$$
Finally, we convert moles of oxygen back to grams using its molar mass:
$$Grams\ of\ O_2 = 1.755\ mol \times 32.00\ g/mol = 56.32\ g$$
Therefore, 56.32 grams of oxygen are needed to react with 23.9 grams of ammonia.
