H2O + HClO3 + P2O3 = HCl + H3PO4

Input interpretation

Image
Text

water + HClO3 + phosphorus trioxide ⟶ hydrogen chloride + phosphoric acid

Balanced equation

Image
Text

$Balance the chemical equation algebraically: + HClO3 + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 HClO3 + c_3 ⟶ c_4 + c_5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Cl and P: H: | 2 c_1 + c_2 = c_4 + 3 c_5 O: | c_1 + 3 c_2 + 3 c_3 = 4 c_5 Cl: | c_2 = c_4 P: | 2 c_3 = c_5 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 9/2 c_2 = 1 c_3 = 3/2 c_4 = 1 c_5 = 3 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 9 c_2 = 2 c_3 = 3 c_4 = 2 c_5 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 + 2 HClO3 + 3 ⟶ 2 + 6$

Balance the chemical equation algebraically: + HClO3 + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 HClO3 + 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, Cl and P: H: | 2 c_1 + c_2 = c_4 + 3 c_5 O: | c_1 + 3 c_2 + 3 c_3 = 4 c_5 Cl: | c_2 = c_4 P: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 9/2 c_2 = 1 c_3 = 3/2 c_4 = 1 c_5 = 3 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 9 c_2 = 2 c_3 = 3 c_4 = 2 c_5 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 + 2 HClO3 + 3 ⟶ 2 + 6