H2O + HNO3 + Sb2O3 = NO + H3SbO4

Input interpretation

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water + nitric acid + antimony trioxide ⟶ nitric oxide + H3SbO4

Balanced equation

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$Balance the chemical equation algebraically: + + ⟶ + H3SbO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 ⟶ c_4 + c_5 H3SbO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and Sb: H: | 2 c_1 + c_2 = 3 c_5 O: | c_1 + 3 c_2 + 3 c_3 = c_4 + 4 c_5 N: | c_2 = c_4 Sb: | 2 c_3 = c_5 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 7/3 c_2 = 4/3 c_3 = 1 c_4 = 4/3 c_5 = 2 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 7 c_2 = 4 c_3 = 3 c_4 = 4 c_5 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 + 4 + 3 ⟶ 4 + 6 H3SbO4$

Balance the chemical equation algebraically: + + ⟶ + H3SbO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 ⟶ c_4 + c_5 H3SbO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and Sb: H: | 2 c_1 + c_2 = 3 c_5 O: | c_1 + 3 c_2 + 3 c_3 = c_4 + 4 c_5 N: | c_2 = c_4 Sb: | 2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 7/3 c_2 = 4/3 c_3 = 1 c_4 = 4/3 c_5 = 2 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 7 c_2 = 4 c_3 = 3 c_4 = 4 c_5 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 + 4 + 3 ⟶ 4 + 6 H3SbO4