Question

Какой объём аммиака (н.у.) выделится при нагревании 21,36г хлорида аммония с избытком гидроксида кальция

Answer 1

2NH4Cl+Ca(OH)2=2NH3+2H2O+CaCl2
∨(NH4Cl)=21,36/53,5=0,4 моль
По уравнению видно, что количества хлорида аммония и аммиака равны:
∨(NH3)=0,4 моль
V(NH3)н.у. = 0,4 (моль) х 22,4 (л/моль)=8,96 л 

Answer 2

Calculation of Ammonia Volume Released

To calculate the volume of ammonia gas released when heating 21.36 grams of ammonium chloride with an excess of calcium hydroxide, we need to use the balanced chemical equation for the reaction between ammonium chloride (NH4Cl) and calcium hydroxide (Ca(OH)2).

The balanced chemical equation for the reaction is as follows:

2NH4Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2O

From the balanced equation, we can see that 2 moles of ammonium chloride react to produce 2 moles of ammonia gas. Therefore, the molar ratio between ammonium chloride and ammonia gas is 1:1.

To calculate the number of moles of ammonium chloride, we need to divide the given mass of ammonium chloride by its molar mass. The molar mass of ammonium chloride (NH4Cl) is calculated as follows:

NH4Cl: - Nitrogen (N) atomic mass: 14.01 g/mol - Hydrogen (H) atomic mass: 1.01 g/mol (x 4 for 4 hydrogen atoms) - Chlorine (Cl) atomic mass: 35.45 g/mol

Molar mass of NH4Cl = (14.01 g/mol) + (1.01 g/mol x 4) + 35.45 g/mol = 53.49 g/mol

Now we can calculate the number of moles of ammonium chloride:

Number of moles of NH4Cl = Mass of NH4Cl / Molar mass of NH4Cl = 21.36 g / 53.49 g/mol = 0.399 moles

Since the molar ratio between ammonium chloride and ammonia gas is 1:1, the number of moles of ammonia gas released will also be 0.399 moles.

To calculate the volume of ammonia gas at standard temperature and pressure (STP), we can use the ideal gas law equation:

PV = nRT

Where: P = Pressure (at STP, pressure = 1 atm) V = Volume n = Number of moles R = Ideal gas constant (0.0821 L·atm/(mol·K)) T = Temperature (at STP, temperature = 273.15 K)

Substituting the values into the equation:

V = (nRT) / P = (0.399 moles) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm) = 8.53 L

Therefore, the volume of ammonia gas released when heating 21.36 grams of ammonium chloride with an excess of calcium hydroxide is approximately 8.53 liters at standard temperature and pressure (STP).