H2O + P + HIO3 = H3PO4 + HI

Input interpretation

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water + red phosphorus + iodic acid ⟶ phosphoric acid + hydrogen iodide

Balanced equation

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$Balance the chemical equation algebraically: + + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 ⟶ c_4 + c_5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and I: H: | 2 c_1 + c_3 = 3 c_4 + c_5 O: | c_1 + 3 c_3 = 4 c_4 P: | c_2 = c_4 I: | c_3 = c_5 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 9/5 c_2 = 6/5 c_3 = 1 c_4 = 6/5 c_5 = 1 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_1 = 9 c_2 = 6 c_3 = 5 c_4 = 6 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 + 6 + 5 ⟶ 6 + 5$

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, P and I: H: | 2 c_1 + c_3 = 3 c_4 + c_5 O: | c_1 + 3 c_3 = 4 c_4 P: | c_2 = c_4 I: | 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 = 9/5 c_2 = 6/5 c_3 = 1 c_4 = 6/5 c_5 = 1 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_1 = 9 c_2 = 6 c_3 = 5 c_4 = 6 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 + 6 + 5 ⟶ 6 + 5