Na + NaNO3 = N2 + Na2O

Na (sodium) + NaNO_3 (sodium nitrate) ⟶ N_2 (nitrogen) + Na_2O (sodium oxide)

Balance the chemical equation algebraically: Na + NaNO_3 ⟶ N_2 + Na_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Na + c_2 NaNO_3 ⟶ c_3 N_2 + c_4 Na_2O Set the number of atoms in the reactants equal to the number of atoms in the products for Na, N and O: Na: | c_1 + c_2 = 2 c_4 N: | c_2 = 2 c_3 O: | 3 c_2 = c_4 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 2 c_3 = 1 c_4 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 Na + 2 NaNO_3 ⟶ N_2 + 6 Na_2O

+ ⟶ +

sodium + sodium nitrate ⟶ nitrogen + sodium oxide

K_c = ([N2] [Na2O]^6)/([Na]^10 [NaNO3]^2)

rate = -1/10 (Δ[Na])/(Δt) = -1/2 (Δ[NaNO3])/(Δt) = (Δ[N2])/(Δt) = 1/6 (Δ[Na2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium | sodium nitrate | nitrogen | sodium oxide formula | Na | NaNO_3 | N_2 | Na_2O Hill formula | Na | NNaO_3 | N_2 | Na_2O name | sodium | sodium nitrate | nitrogen | sodium oxide IUPAC name | sodium | sodium nitrate | molecular nitrogen | disodium oxygen(-2) anion

| sodium | sodium nitrate | nitrogen | sodium oxide molar mass | 22.98976928 g/mol | 84.994 g/mol | 28.014 g/mol | 61.979 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | melting point | 97.8 °C | 306 °C | -210 °C | boiling point | 883 °C | | -195.79 °C | density | 0.968 g/cm^3 | 2.26 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) | 2.27 g/cm^3 solubility in water | decomposes | soluble | insoluble | surface tension | | | 0.0066 N/m | dynamic viscosity | 1.413×10^-5 Pa s (at 527 °C) | 0.003 Pa s (at 250 °C) | 1.78×10^-5 Pa s (at 25 °C) | odor | | | odorless |