Question

Вычислите объём аммиака (н.у) который выделится при взаимодействии 890г нитрата аммония,содержащего 20% примесей с раствором гидроксида натрия .Сколько граммов 20%го раствора щелочи потребуется для реакции?(пожалуйста как можно подробней)

Answer 1

NH4NO3+NaOH=NH3(стрелочка вверх)+NaNO3+H2O

m(NH4NO3)=890 г

w(примесей)=20%

w(р-ра)=20%

 m(NaOH)-?

 

Для начала найдем массу чистого нитрата аммония

w=m(в-ва)/m(р-ра)*100%

0,2= m(в-ва)/890, тогда m(в-ва)=0,2*890=178 г

Из реакции:
n(NH4NO3) = 178 / (14*2+4+16*3) = 2,225 моль = n(NaOH) 

Тогда m(NaOH)=2,225*(23+16+1) = 89 г

 w=m(в-ва)/m(р-ра)*100%

0,2=89/m(р-ра)

m(р-ра)=89/0,2=445 г

 

Ответ : m(р-ра)=445 г

Answer 2

Calculation of Ammonia Volume

To calculate the volume of ammonia gas released during the reaction, we need to determine the amount of ammonium nitrate and the stoichiometry of the reaction.

Given: - Mass of ammonium nitrate (NH4NO3) = 890 g - Ammonium nitrate contains 20% impurities - Reaction with sodium hydroxide (NaOH) produces ammonia gas (NH3)

First, let's calculate the mass of pure ammonium nitrate in the given sample: - Mass of pure ammonium nitrate = Mass of ammonium nitrate - Mass of impurities - Mass of impurities = 20% of mass of ammonium nitrate

Let's calculate the mass of impurities: - Mass of impurities = 20/100 * 890 g

Now, subtract the mass of impurities from the mass of ammonium nitrate to get the mass of pure ammonium nitrate: - Mass of pure ammonium nitrate = 890 g - (20/100 * 890 g)

Next, we need to determine the stoichiometry of the reaction between ammonium nitrate and sodium hydroxide. The balanced equation for the reaction is:

2NH4NO3 + 2NaOH → 2NH3 + 2H2O + Na2CO3

From the balanced equation, we can see that 2 moles of ammonium nitrate produce 2 moles of ammonia gas. Therefore, the stoichiometric ratio is 1:1.

Now, let's calculate the number of moles of pure ammonium nitrate: - Moles of pure ammonium nitrate = Mass of pure ammonium nitrate / Molar mass of ammonium nitrate

The molar mass of ammonium nitrate (NH4NO3) can be calculated as follows: - Molar mass of NH4NO3 = (Molar mass of N) + 4*(Molar mass of H) + (Molar mass of N) + 3*(Molar mass of O)

Finally, we can calculate the volume of ammonia gas using the ideal gas law: - Volume of ammonia gas = Moles of pure ammonium nitrate * 22.4 L/mol (at standard temperature and pressure)

Calculation of Sodium Hydroxide Solution

To determine the amount of 20% sodium hydroxide solution required for the reaction, we need to consider the stoichiometry of the reaction and the concentration of the sodium hydroxide solution.

Given: - Concentration of sodium hydroxide solution = 20% - Balanced equation: 2NH4NO3 + 2NaOH → 2NH3 + 2H2O + Na2CO3

From the balanced equation, we can see that 2 moles of sodium hydroxide react with 2 moles of ammonium nitrate to produce 2 moles of ammonia gas.

To calculate the amount of sodium hydroxide solution required, we need to determine the number of moles of ammonium nitrate. We already calculated this value earlier.

Now, let's calculate the number of moles of sodium hydroxide required: - Moles of sodium hydroxide = Moles of pure ammonium nitrate

Finally, we can calculate the mass of 20% sodium hydroxide solution required using the concentration and molar mass of sodium hydroxide: - Mass of 20% sodium hydroxide solution = Moles of sodium hydroxide * Molar mass of NaOH

Please note that the molar mass of NaOH can be calculated as follows: - Molar mass of NaOH = (Molar mass of Na) + (Molar mass of O) + (Molar mass of H)

Please note that the exact calculation may require additional information, such as the molar masses of the elements involved.

Let me know if you need any further assistance!