Na + F2 = NaF

sodium + fluorine ⟶ sodium fluoride

Balance the chemical equation algebraically: + ⟶ Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Na and F: Na: | c_1 = c_3 F: | 2 c_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 = 2 c_2 = 1 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + ⟶ 2

+ ⟶

sodium + fluorine ⟶ sodium fluoride

| sodium | fluorine | sodium fluoride molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -576.6 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -1153 kJ/mol | H_initial = 0 kJ/mol | | H_final = -1153 kJ/mol ΔH_rxn^0 | -1153 kJ/mol - 0 kJ/mol = -1153 kJ/mol (exothermic) | |

| sodium | fluorine | sodium fluoride molecular entropy | 51 J/(mol K) | 202.8 J/(mol K) | 51.1 J/(mol K) total entropy | 102 J/(mol K) | 202.8 J/(mol K) | 102.2 J/(mol K) | S_initial = 304.8 J/(mol K) | | S_final = 102.2 J/(mol K) ΔS_rxn^0 | 102.2 J/(mol K) - 304.8 J/(mol K) = -202.6 J/(mol K) (exoentropic) | |

| sodium | fluorine | sodium fluoride Hill formula | Na | F_2 | FNa name | sodium | fluorine | sodium fluoride IUPAC name | sodium | molecular fluorine | sodium fluoride

| sodium | fluorine | sodium fluoride molar mass | 22.98976928 g/mol | 37.996806326 g/mol | 41.98817244 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 97.8 °C | -219.6 °C | 993 °C boiling point | 883 °C | -188.12 °C | 1700 °C density | 0.968 g/cm^3 | 0.001696 g/cm^3 (at 0 °C) | 2.558 g/cm^3 solubility in water | decomposes | reacts | dynamic viscosity | 1.413×10^-5 Pa s (at 527 °C) | 2.344×10^-5 Pa s (at 25 °C) | 0.00105 Pa s (at 1160 °C) odor | | | odorless