HNO3 + Cd = H2O + NO + Cd(NO3)2

nitric acid + cadmium ⟶ water + nitric oxide + Cd(NO3)2

Balance the chemical equation algebraically: + ⟶ + + Cd(NO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 + c_5 Cd(NO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Cd: H: | c_1 = 2 c_3 N: | c_1 = c_4 + 2 c_5 O: | 3 c_1 = c_3 + c_4 + 6 c_5 Cd: | 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 = 4 c_2 = 3/2 c_3 = 2 c_4 = 1 c_5 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 3 c_3 = 4 c_4 = 2 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 + 3 ⟶ 4 + 2 + 3 Cd(NO3)2

+ ⟶ + + Cd(NO3)2

nitric acid + cadmium ⟶ water + nitric oxide + Cd(NO3)2

| nitric acid | cadmium | water | nitric oxide | Cd(NO3)2 formula | | | | | Cd(NO3)2 Hill formula | HNO_3 | Cd | H_2O | NO | CdN2O6 name | nitric acid | cadmium | water | nitric oxide |

| nitric acid | cadmium | water | nitric oxide | Cd(NO3)2 molar mass | 63.012 g/mol | 112.414 g/mol | 18.015 g/mol | 30.006 g/mol | 236.42 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 320.9 °C | 0 °C | -163.6 °C | boiling point | 83 °C | 765 °C | 99.9839 °C | -151.7 °C | density | 1.5129 g/cm^3 | 8.65 g/cm^3 | 1 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.911×10^-5 Pa s (at 25 °C) | odor | | odorless | odorless | |