H2O + Br2 + P = H3PO4 + HBr

Input interpretation

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water + bromine + red phosphorus ⟶ phosphoric acid + hydrogen bromide

Balanced equation

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$Balance the chemical equation algebraically: + + ⟶ + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 ⟶ c_4 + c_5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and P: H: | 2 c_1 = 3 c_4 + c_5 O: | c_1 = 4 c_4 Br: | 2 c_2 = c_5 P: | c_3 = c_4 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 5/2 c_3 = 1 c_4 = 1 c_5 = 5 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 5 c_3 = 2 c_4 = 2 c_5 = 10 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 + 5 + 2 ⟶ 2 + 10$

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, Br and P: H: | 2 c_1 = 3 c_4 + c_5 O: | c_1 = 4 c_4 Br: | 2 c_2 = c_5 P: | c_3 = 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 = 4 c_2 = 5/2 c_3 = 1 c_4 = 1 c_5 = 5 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 5 c_3 = 2 c_4 = 2 c_5 = 10 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 + 5 + 2 ⟶ 2 + 10