H2O + NaOH + Al = H2 + NaAlO2

Input interpretation

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water + sodium hydroxide + aluminum ⟶ hydrogen + sodium aluminate

Balanced equation

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

Balance the chemical equation algebraically: + + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 ⟶ c_4 + c_5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Na and Al: H: | 2 c_1 + c_2 = 2 c_4 O: | c_1 + c_2 = 2 c_5 Na: | c_2 = c_5 Al: | c_3 = c_5 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 3/2 c_5 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 2 c_3 = 2 c_4 = 3 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + 2 + 2 ⟶ 3 + 2