H2O + BrF3 = HF + HBrO2

water + bromine trifluoride ⟶ hydrogen fluoride + HBrO2

Balance the chemical equation algebraically: + ⟶ + HBrO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 HBrO2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and F: H: | 2 c_1 = c_3 + c_4 O: | c_1 = 2 c_4 Br: | c_2 = c_4 F: | 3 c_2 = c_3 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 = 2 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 + ⟶ 3 + HBrO2

+ ⟶ + HBrO2

water + bromine trifluoride ⟶ hydrogen fluoride + HBrO2

| water | bromine trifluoride | hydrogen fluoride | HBrO2 formula | | | | HBrO2 Hill formula | H_2O | BrF_3 | FH | HBrO2 name | water | bromine trifluoride | hydrogen fluoride |

| water | bromine trifluoride | hydrogen fluoride | HBrO2 molar mass | 18.015 g/mol | 136.9 g/mol | 20.006 g/mol | 112.91 g/mol phase | liquid (at STP) | liquid (at STP) | gas (at STP) | melting point | 0 °C | 8.77 °C | -83.36 °C | boiling point | 99.9839 °C | 125.8 °C | 19.5 °C | density | 1 g/cm^3 | 2.803 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | solubility in water | | decomposes | miscible | surface tension | 0.0728 N/m | 0.0363 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 1.2571×10^-5 Pa s (at 20 °C) | odor | odorless | | |