Question

Вычислите объем аммиака (н.у.) который выделится при взаимодействии 890 г нитрата амония содержащего 20% примисей с раствором гидроксида натрия Сколько граммов 20% раствора щелочи потребуется для реакции?

Answer 1

NH4NO3+NaOH=NH3+NaNO3+H2
чистая масса нитрата аммония=890*0,8=712 г
кол-во нитрата аммония=712/80=8,9 моль
масса грязная щёлочи=8,9*40=356 г.
масса чистая щёлочи=356*0,2=71,2 г.
объём выделившегося аммиака=8,9*22,4=199,36 л.
Задача не понятная,проверяйте решение,могут быть ошибки

Answer 2

Calculation of Ammonia Volume:

To calculate the volume of ammonia gas (at standard conditions) that will be released during the reaction, we need to determine the moles of ammonium nitrate and then use the stoichiometry of the reaction to find the moles of ammonia gas produced.

Given: - Mass of ammonium nitrate (NH4NO3) = 890 g - Ammonium nitrate contains 20% impurities - We need to find the volume of ammonia gas released

To calculate the moles of ammonium nitrate, we need to subtract the mass of impurities from the total mass of ammonium nitrate: - Mass of pure ammonium nitrate = 890 g - (20% of 890 g) = 890 g - 178 g = 712 g

Next, we calculate the moles of ammonium nitrate using its molar mass: - Molar mass of ammonium nitrate (NH4NO3) = 80.043 g/mol - Moles of ammonium nitrate = Mass / Molar mass = 712 g / 80.043 g/mol = 8.893 mol

The balanced chemical equation for the reaction between ammonium nitrate and sodium hydroxide is: 2 NH4NO3 + 2 NaOH → Na2CO3 + 2 H2O + 2 NH3

From the balanced equation, we can see that 2 moles of ammonium nitrate produce 2 moles of ammonia gas. Therefore, the moles of ammonia gas produced will be equal to the moles of ammonium nitrate: - Moles of ammonia gas = 8.893 mol

To calculate the volume of ammonia gas at standard conditions (n.у.), we can use the ideal gas law: - V = n * Vm

Where: - V is the volume of the gas - n is the number of moles of the gas - Vm is the molar volume of the gas at standard conditions (22.4 L/mol)

Substituting the values: - V = 8.893 mol * 22.4 L/mol = 199.0512 L

Therefore, the volume of ammonia gas (at standard conditions) that will be released during the reaction is approximately 199.05 liters.

Calculation of Sodium Hydroxide Solution:

To calculate the amount of a 20% sodium hydroxide solution required for the reaction, we need to determine the moles of sodium hydroxide and then use the stoichiometry of the reaction to find the moles of sodium hydroxide consumed.

Given: - We need to find the mass of a 20% sodium hydroxide solution required

To calculate the moles of sodium hydroxide, we need to determine the mass of pure sodium hydroxide in the solution: - Mass of sodium hydroxide in the solution = 20% of the mass of the solution

Let's assume the mass of the 20% sodium hydroxide solution required is x grams. Then: - Mass of sodium hydroxide in the solution = 20% of x grams = 0.2x grams

Next, we calculate the moles of sodium hydroxide using its molar mass: - Molar mass of sodium hydroxide (NaOH) = 39.997 g/mol - Moles of sodium hydroxide = Mass / Molar mass = (0.2x grams) / 39.997 g/mol

From the balanced chemical equation, we can see that 2 moles of sodium hydroxide react with 2 moles of ammonium nitrate. Therefore, the moles of sodium hydroxide consumed will be equal to the moles of ammonium nitrate: - Moles of sodium hydroxide = 8.893 mol

Setting up the equation: (0.2x grams) / 39.997 g/mol = 8.893 mol

Solving for x: x = (8.893 mol * 39.997 g/mol) / 0.2

Therefore, the mass of the 20% sodium hydroxide solution required for the reaction is approximately 889.1 grams.

Please note that the calculations provided are based on the information given and assumptions made. It is always recommended to double-check the calculations and use accurate experimental data for precise results.