O2 + P4 = P2O5

oxygen + white phosphorus ⟶ P2O5

Balance the chemical equation algebraically: + ⟶ P2O5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 P2O5 Set the number of atoms in the reactants equal to the number of atoms in the products for O and P: O: | 2 c_1 = 5 c_3 P: | 4 c_2 = 2 c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5 c_2 = 1 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 + ⟶ 2 P2O5

+ ⟶ P2O5

oxygen + white phosphorus ⟶ P2O5

| oxygen | white phosphorus | P2O5 formula | | | P2O5 Hill formula | O_2 | P_4 | O5P2 name | oxygen | white phosphorus | IUPAC name | molecular oxygen | tetraphosphorus |

| oxygen | white phosphorus | P2O5 molar mass | 31.998 g/mol | 123.89504799 g/mol | 141.94 g/mol phase | gas (at STP) | solid (at STP) | melting point | -218 °C | 44.15 °C | boiling point | -183 °C | 280.5 °C | density | 0.001429 g/cm^3 (at 0 °C) | 1.823 g/cm^3 | solubility in water | | insoluble | surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 0.00169 Pa s (at 50 °C) | odor | odorless | odorless |