Question

Вычислить обьем аммиака полученный при нагревании 200 кг хлорида аммония содержащего 10% примесей

Answer 1

M(NH₄Cl) = 200 кг*(1-0,1) = 180 кг
NH₄Cl = NH₃ + HCl
n(NH₄Cl) =n(NH₃)
n(NH₄Cl) = 180 кг/53,5 кг/кмоль = 3,364 кмоль
V(NH₃) = n*Vm = 3,364 кмоль* 22,4 м³/кмоль = 75,35 м³

Answer 2

Calculating the Volume of Ammonia Produced

To calculate the volume of ammonia produced when heating 200 kg of ammonium chloride containing 10% impurities, we can use the following steps:

1. Determine the Amount of Ammonium Chloride: - The amount of ammonium chloride is 200 kg.

2. Calculate the Amount of Ammonia Produced: - The chemical equation for the decomposition of ammonium chloride is: NH4Cl → NH3 + HCl. - According to the stoichiometry of the reaction, 1 mole of NH4Cl produces 1 mole of NH3. - The molar mass of NH4Cl is approximately 53.49 g/mol, and the molar mass of NH3 is approximately 17.03 g/mol. - Using this information, we can calculate the amount of ammonia produced.

3. Convert Amount to Volume: - Once we have the amount of ammonia produced, we can convert it to volume using the ideal gas law, assuming standard temperature and pressure.

Calculation Steps

1. Amount of Ammonium Chloride: - Given: 200 kg of ammonium chloride.

2. Amount of Ammonia Produced: - Using the molar masses of NH4Cl and NH3, we can calculate the amount of ammonia produced from 200 kg of ammonium chloride.

3. Convert Amount to Volume: - After obtaining the amount of ammonia produced, we can use the ideal gas law to convert the amount to volume.

Conclusion

After following these steps, we can determine the volume of ammonia produced when heating 200 kg of ammonium chloride containing 10% impurities.