CaO + SO3 = CaSO4

CaO lime + SO_3 sulfur trioxide ⟶ CaSO_4 calcium sulfate

Balance the chemical equation algebraically: CaO + SO_3 ⟶ CaSO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CaO + c_2 SO_3 ⟶ c_3 CaSO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Ca, O and S: Ca: | c_1 = c_3 O: | c_1 + 3 c_2 = 4 c_3 S: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | CaO + SO_3 ⟶ CaSO_4

+ ⟶

lime + sulfur trioxide ⟶ calcium sulfate

K_c = [CaSO4]/([CaO] [SO3])

rate = -(Δ[CaO])/(Δt) = -(Δ[SO3])/(Δt) = (Δ[CaSO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| lime | sulfur trioxide | calcium sulfate formula | CaO | SO_3 | CaSO_4 Hill formula | CaO | O_3S | CaO_4S name | lime | sulfur trioxide | calcium sulfate

| lime | sulfur trioxide | calcium sulfate molar mass | 56.077 g/mol | 80.06 g/mol | 136.13 g/mol phase | solid (at STP) | liquid (at STP) | melting point | 2580 °C | 16.8 °C | boiling point | 2850 °C | 44.7 °C | density | 3.3 g/cm^3 | 1.97 g/cm^3 | solubility in water | reacts | reacts | slightly soluble dynamic viscosity | | 0.00159 Pa s (at 30 °C) | odor | | | odorless