Before solving problems, it is necessary to consider formulas andThe rules of how to find the volume of gas. It should be remembered the law of Avogadro. And the volume of gas itself can be calculated with the aid of several formulas, choosing from them a suitable one. When choosing the necessary formula, environmental conditions, in particular temperature and pressure, are of great importance.

Avogadro's Law

It says that at the same pressure andthe same temperature, in the same volumes of different gases, will contain the same number of molecules. The number of gas molecules contained in one mole is the Avogadro number. From this law it follows that: 1 kmol of ideal gas, and any one, at the same pressure and temperature (760 mm Hg and t = 0 * C) always occupies one volume = 22,4136 m3.

How to determine the volume of gas

  • The formula V = n * Vm is most often found intasks. Here the volume of gas in liters is V, Vm is the volume of gas molar (l / mol), which under normal conditions is 22.4 l / mol, and n is the amount of matter in moles. When there is no quantity of substance in the conditions, but there is a mass of matter, then we proceed in this way: n = m / M. Here, M is g / mol (the molar mass of the substance), and the mass of the substance in grams is m. In the periodic table, it is written under each element, as its atomic mass. We add all the masses and obtain the required one.
  • So, how to calculate the volume of gas. Here's the problem: dissolve 10 grams of aluminum in hydrochloric acid. Question: how much hydrogen can stand out when n. y.? The reaction equation is as follows: 2Al + 6HCl (g) = 2AlCl3 + 3H2. At the very beginning, we find aluminum (quantity), reacted according to the formula: n (Al) = m (Al) / M (Al). We take the mass of aluminum (molar) from the Mendeleyev table M (Al) = 27 g / mol. We substitute: n (Al) = 10/27 = 0.37 mol. It is clear from the chemical equation that 3 moles of hydrogen formed when two moles of aluminum were dissolved. It should be calculated, and how much hydrogen will be released from 0.4 moles of aluminum: n (H2) = 3 * 0.37 / 2 = 0.56 mol. We substitute the data in the formula and find the volume of this gas. V = n * Vm = 0.56 * 22.4 = 12.54L.
Comments 0