O2 + B5H9 = H2O + B2O3

O_2 oxygen + B_5H_9 pentaborane(9) ⟶ H_2O water + B_2O_3 boron oxide

Balance the chemical equation algebraically: O_2 + B_5H_9 ⟶ H_2O + B_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 B_5H_9 ⟶ c_3 H_2O + c_4 B_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O, B and H: O: | 2 c_1 = c_3 + 3 c_4 B: | 5 c_2 = 2 c_4 H: | 9 c_2 = 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 = 6 c_2 = 1 c_3 = 9/2 c_4 = 5/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 12 c_2 = 2 c_3 = 9 c_4 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 O_2 + 2 B_5H_9 ⟶ 9 H_2O + 5 B_2O_3

+ ⟶ +

oxygen + pentaborane(9) ⟶ water + boron oxide

K_c = ([H2O]^9 [B2O3]^5)/([O2]^12 [B5H9]^2)

rate = -1/12 (Δ[O2])/(Δt) = -1/2 (Δ[B5H9])/(Δt) = 1/9 (Δ[H2O])/(Δt) = 1/5 (Δ[B2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | pentaborane(9) | water | boron oxide formula | O_2 | B_5H_9 | H_2O | B_2O_3 name | oxygen | pentaborane(9) | water | boron oxide IUPAC name | molecular oxygen | | water |

| oxygen | pentaborane(9) | water | boron oxide molar mass | 31.998 g/mol | 63.12 g/mol | 18.015 g/mol | 69.62 g/mol phase | gas (at STP) | | liquid (at STP) | solid (at STP) melting point | -218 °C | -46.74 °C | 0 °C | 450 °C boiling point | -183 °C | 60.1 °C | 99.9839 °C | 1860 °C density | 0.001429 g/cm^3 (at 0 °C) | 0.6 g/cm^3 | 1 g/cm^3 | 2.46 g/cm^3 surface tension | 0.01347 N/m | | 0.0728 N/m | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 85 Pa s (at 700 °C) odor | odorless | | odorless |