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NaOH + H2CO3 = H2O + Na2CO3

Input interpretation

sodium hydroxide + carbonic acid ⟶ water + soda ash
sodium hydroxide + carbonic acid ⟶ water + soda ash

Balanced equation

Balance the chemical equation algebraically:  + ⟶ +  Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4  Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O and C: H: | c_1 + 2 c_2 = 2 c_3 Na: | c_1 = 2 c_4 O: | c_1 + 3 c_2 = c_3 + 3 c_4 C: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 2 + ⟶ 2 +
Balance the chemical equation algebraically: + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O and C: H: | c_1 + 2 c_2 = 2 c_3 Na: | c_1 = 2 c_4 O: | c_1 + 3 c_2 = c_3 + 3 c_4 C: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + ⟶ 2 +

Structures

 + ⟶ +
+ ⟶ +

Names

sodium hydroxide + carbonic acid ⟶ water + soda ash
sodium hydroxide + carbonic acid ⟶ water + soda ash