I2 + NaHCO3 + As = H2O + CO2 + NaI + Na2HAsO4

Input interpretation

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I_2 iodine + NaHCO_3 sodium bicarbonate + As gray arsenic ⟶ H_2O water + CO_2 carbon dioxide + NaI sodium iodide + AsHNa_2O_4 disodium hydrogen arsenate

Balanced equation

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$Balance the chemical equation algebraically: I_2 + NaHCO_3 + As ⟶ H_2O + CO_2 + NaI + AsHNa_2O_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 I_2 + c_2 NaHCO_3 + c_3 As ⟶ c_4 H_2O + c_5 CO_2 + c_6 NaI + c_7 AsHNa_2O_4 Set the number of atoms in the reactants equal to the number of atoms in the products for I, C, H, Na, O and As: I: | 2 c_1 = c_6 C: | c_2 = c_5 H: | c_2 = 2 c_4 + c_7 Na: | c_2 = c_6 + 2 c_7 O: | 3 c_2 = c_4 + 2 c_5 + 4 c_7 As: | c_3 = c_7 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5/2 c_2 = 7 c_3 = 1 c_4 = 3 c_5 = 7 c_6 = 5 c_7 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 5 c_2 = 14 c_3 = 2 c_4 = 6 c_5 = 14 c_6 = 10 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 I_2 + 14 NaHCO_3 + 2 As ⟶ 6 H_2O + 14 CO_2 + 10 NaI + 2 AsHNa_2O_4$

Balance the chemical equation algebraically: I_2 + NaHCO_3 + As ⟶ H_2O + CO_2 + NaI + AsHNa_2O_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 I_2 + c_2 NaHCO_3 + c_3 As ⟶ c_4 H_2O + c_5 CO_2 + c_6 NaI + c_7 AsHNa_2O_4 Set the number of atoms in the reactants equal to the number of atoms in the products for I, C, H, Na, O and As: I: | 2 c_1 = c_6 C: | c_2 = c_5 H: | c_2 = 2 c_4 + c_7 Na: | c_2 = c_6 + 2 c_7 O: | 3 c_2 = c_4 + 2 c_5 + 4 c_7 As: | c_3 = c_7 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 = 5/2 c_2 = 7 c_3 = 1 c_4 = 3 c_5 = 7 c_6 = 5 c_7 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 5 c_2 = 14 c_3 = 2 c_4 = 6 c_5 = 14 c_6 = 10 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 I_2 + 14 NaHCO_3 + 2 As ⟶ 6 H_2O + 14 CO_2 + 10 NaI + 2 AsHNa_2O_4

+ + ⟶ + + +

Names

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iodine + sodium bicarbonate + gray arsenic ⟶ water + carbon dioxide + sodium iodide + disodium hydrogen arsenate

Equilibrium constant

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K_c = ([H2O]^6 [CO2]^14 [NaI]^10 [AsHNa2O4]^2)/([I2]^5 [NaHCO3]^14 [As]^2)

Rate of reaction

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rate = -1/5 (Δ[I2])/(Δt) = -1/14 (Δ[NaHCO3])/(Δt) = -1/2 (Δ[As])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/14 (Δ[CO2])/(Δt) = 1/10 (Δ[NaI])/(Δt) = 1/2 (Δ[AsHNa2O4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)