KOH + (NH4)2SO4 = H2O + K2SO4 + NH3

KOH (potassium hydroxide) + (NH_4)_2SO_4 (ammonium sulfate) ⟶ H_2O (water) + K_2SO_4 (potassium sulfate) + NH_3 (ammonia)

Balance the chemical equation algebraically: KOH + (NH_4)_2SO_4 ⟶ H_2O + K_2SO_4 + NH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 (NH_4)_2SO_4 ⟶ c_3 H_2O + c_4 K_2SO_4 + c_5 NH_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, N and S: H: | c_1 + 8 c_2 = 2 c_3 + 3 c_5 K: | c_1 = 2 c_4 O: | c_1 + 4 c_2 = c_3 + 4 c_4 N: | 2 c_2 = c_5 S: | c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 2 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + (NH_4)_2SO_4 ⟶ 2 H_2O + K_2SO_4 + 2 NH_3

+ ⟶ + +

potassium hydroxide + ammonium sulfate ⟶ water + potassium sulfate + ammonia

K_c = ([H2O]^2 [K2SO4] [NH3]^2)/([KOH]^2 [(NH4)2SO4])

rate = -1/2 (Δ[KOH])/(Δt) = -(Δ[(NH4)2SO4])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[K2SO4])/(Δt) = 1/2 (Δ[NH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | ammonium sulfate | water | potassium sulfate | ammonia formula | KOH | (NH_4)_2SO_4 | H_2O | K_2SO_4 | NH_3 Hill formula | HKO | H_8N_2O_4S | H_2O | K_2O_4S | H_3N name | potassium hydroxide | ammonium sulfate | water | potassium sulfate | ammonia IUPAC name | potassium hydroxide | | water | dipotassium sulfate | ammonia