HNO3 + PbO2 + MoS = H2O + H2SO4 + Pb(NO3)2 + H2MoO4

Input interpretation

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nitric acid + lead dioxide + MoS ⟶ water + sulfuric acid + lead(II) nitrate + molybdic acid

Balanced equation

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Balance the chemical equation algebraically: + + MoS ⟶ + + + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 + c_3 MoS ⟶ c_4 + c_5 + c_6 + c_7 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Pb, Mo and S: H: | c_1 = 2 c_4 + 2 c_5 + 2 c_7 N: | c_1 = 2 c_6 O: | 3 c_1 + 2 c_2 = c_4 + 4 c_5 + 6 c_6 + 4 c_7 Pb: | c_2 = c_6 Mo: | c_3 = c_7 S: | c_3 = c_5 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 12 c_2 = 6 c_3 = 1 c_4 = 4 c_5 = 1 c_6 = 6 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 + 6 + MoS ⟶ 4 + + 6 +