HNO3 + Sn = H2O + N2O + Sn(NO3)2

nitric acid + white tin ⟶ water + nitrous oxide + Sn(NO3)2

Balance the chemical equation algebraically: + ⟶ + + Sn(NO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 + c_5 Sn(NO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Sn: H: | c_1 = 2 c_3 N: | c_1 = 2 c_4 + 2 c_5 O: | 3 c_1 = c_3 + c_4 + 6 c_5 Sn: | c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 4 c_3 = 5 c_4 = 1 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 + 4 ⟶ 5 + + 4 Sn(NO3)2

+ ⟶ + + Sn(NO3)2

nitric acid + white tin ⟶ water + nitrous oxide + Sn(NO3)2

| nitric acid | white tin | water | nitrous oxide | Sn(NO3)2 formula | | | | | Sn(NO3)2 Hill formula | HNO_3 | Sn | H_2O | N_2O | N2O6Sn name | nitric acid | white tin | water | nitrous oxide | IUPAC name | nitric acid | tin | water | nitrous oxide |