This reaction does take place, but the hydrogen set free is immediately oxidized to water by another portion of the nitric acid according to the equation HNO_{3} + 3H = 2H_{2}O + NO. As these two equations are written, two atoms of hydrogen are given off in the first equation, while three are used up in the second. In order that the hydrogen may be equal in the two equations, we must multiply the first by 3 and the second by 2. We shall then have 3Cu + 6HNO_{3} = 3Cu(NO_{3})_{2} + 6H, 2HNO_{3} + 6H = 4H_{2}O + 2NO. The two equations may now be combined into one by adding the quantities on each side of the equality sign, canceling the hydrogen which is given off in the one reaction and used up in the other. We shall then have the equation 3Cu + 8HNO_{3} = 3Cu(NO_{3})_{2} + 2NO + 4H_{2}O. A number of other reactions may take place when nitric acid acts upon metals, resulting in the formation of other oxides of nitrogen, free nitrogen, or even ammonia. The reaction just given is, however, the usual one. ~Importance of steps in a reaction.~ This complete equation has the advantage of making it possible to calculate very easily the proportions in which the various substances enter into the reaction or are formed in it. It is unsatisfactory in that it does not give full information about the way in which the reaction takes place. For example, it does not suggest that hydrogen is at first formed, and subsequently transformed into water. It is always much more important to remember the steps in a chemical reaction than to remember the equation expressing the complete action; for if these steps in the reaction are understood, the complete equation is easily obtained in the manner just described. ~Salts of nitric acid,--nitrates.~ The salts of nitric acid are called nitrates. Many of these salts will be described in the study of the metals. They are all soluble in water, and when heated to a high temperature undergo decomposition. In a few cases a nitrate on being heated evolves oxygen, forming a nitrite: NaNO_{3} = NaNO_{2} + O. In other cases the decomposition goes further, and the metal is left as oxide: Cu(NO_{3})_{2} = CuO + 2NO_{2} + O. ~Nitrous acid~ (HNO_{2}). It is an easy matter to obtain sodium nitrite (NaNO_{2}), as the reaction given on the previous page indicates. Instead of merely heating the nitrate, it is better to heat it together with a mild reducing agent, such as lead, when the reaction takes place which is expressed by the equation NaNO_{3} + Pb = PbO + NaNO_{2}. When sodium nitrite is treated with an acid, such as sulphuric acid, it is decomposed and nitrous acid is set free: NaNO_{2} + H_{2}SO_{4} = NaHSO_{4} + HNO_{2}. The acid is very unstable, however, and decomposes readily into water and nitrogen trioxide (N_{2}O_{3}): 2HNO_{2} = H_{2}O + N_{2}O_{3}. Dilute solutions of the acid, however, can be obtained.