Chapter 3 - Equations, the Mole, and Chemical Formulas - Questions and Exercises - Exercises - Page 133: 3.48

$P_{4} (s) + 5O_{2} (g) → P_{4}O_{10}$ Oxidation states (reactants): P = 0, O = 0 Oxidation states (products): P = +5, O = -2 Hence, oxygen is reduced and phosphorus is oxidized.

$P_{4} (s) + 5O_{2} (g) → P_{4}O_{10}$ Oxidation states (reactants): P = 0, O = 0 Oxidation states (products): P = +5, O = -2 Hence, oxygen is reduced and phosphorus is oxidized.