H2SO4 + Ba(NO3)2 = HNO3 + BaSO4

H_2SO_4 (sulfuric acid) + Ba(NO_3)_2 (barium nitrate) ⟶ HNO_3 (nitric acid) + BaSO_4 (barium sulfate)

Balance the chemical equation algebraically: H_2SO_4 + Ba(NO_3)_2 ⟶ HNO_3 + BaSO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2SO_4 + c_2 Ba(NO_3)_2 ⟶ c_3 HNO_3 + c_4 BaSO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, S, Ba and N: H: | 2 c_1 = c_3 O: | 4 c_1 + 6 c_2 = 3 c_3 + 4 c_4 S: | c_1 = c_4 Ba: | c_2 = c_4 N: | 2 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 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2SO_4 + Ba(NO_3)_2 ⟶ 2 HNO_3 + BaSO_4