HNO3 + S = H2O + SO2 + NO

Input interpretation

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nitric acid + mixed sulfur ⟶ water + sulfur dioxide + nitric oxide

Balanced equation

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

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, N, O and S: H: | c_1 = 2 c_3 N: | c_1 = c_5 O: | 3 c_1 = c_3 + 2 c_4 + c_5 S: | 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 = 2 c_2 = 3/2 c_3 = 1 c_4 = 3/2 c_5 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 3 c_3 = 2 c_4 = 3 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 + 3 ⟶ 2 + 3 + 4