Question

При взаимодействии 67,2 л азота с водородом образовалось 33,6 л аммиака .Узнать массовую долю (по объему) практического выхода аммиака от теоретического

Answer 1

Подробное решение в приложении 

Answer 2

Calculation of the Mass Fraction of Ammonia

To calculate the mass fraction of ammonia formed when 67.2 L of nitrogen reacts with hydrogen to produce 33.6 L of ammonia, we need to use the ideal gas law and the molar masses of the gases involved.

The balanced chemical equation for the reaction is:

N2 + 3H2 → 2NH3

From the equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

To find the number of moles of nitrogen, hydrogen, and ammonia, we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure of the gas (assumed constant) - V is the volume of the gas - n is the number of moles of the gas - R is the ideal gas constant - T is the temperature in Kelvin (assumed constant)

Since the pressure, temperature, and volume are constant, we can simplify the equation to:

n = V / Vm

Where: - V is the volume of the gas - Vm is the molar volume of the gas at the given conditions (standard temperature and pressure)

The molar volume of an ideal gas at standard temperature and pressure (STP) is 22.4 L/mol.

Using this information, we can calculate the number of moles of nitrogen, hydrogen, and ammonia:

- Moles of nitrogen (N2) = 67.2 L / 22.4 L/mol = 3 moles - Moles of hydrogen (H2) = 3 * 67.2 L / 22.4 L/mol = 9 moles - Moles of ammonia (NH3) = 2 * 33.6 L / 22.4 L/mol = 3 moles

Now, we can calculate the mass fraction of ammonia:

Mass fraction of ammonia = (mass of ammonia) / (total mass of products)

The molar mass of ammonia (NH3) is 17.03 g/mol.

- Mass of ammonia = 3 moles * 17.03 g/mol = 51.09 g - Total mass of products = mass of ammonia

Therefore, the mass fraction of ammonia is:

Mass fraction of ammonia = 51.09 g / 51.09 g = 1

The mass fraction of ammonia is 1, which means that the practical yield of ammonia is equal to the theoretical yield.