Question

Нитрит калия массой 8,5г внесли при нагревании в 270г раствора бромида аммония с массовой долей 12%. Какой объем(н.у.) азота выделится при этом и какова массовая доля бромида аммония в получившемся растворе? помогите пожалуйста

Answer 1

КNO₂ + NH₄Br = N₂↑ + KBr + 2H₂O
n(KNO₂) = n(NH₄Br) =n(N₂)
m₁(NH₄Br) = ω*m(p.NH₄Br) =0,12*270 г.= 32,4 г.
n₁(NH₄Br) = m/M = 32,4 г.: 98 г/моль = 0,3306 моль
n(KNO₂) = m/M = 8,5г/85 г/моль = 0,1 моль -- считаем по веществу, взятому в недостатке  ⇒ n(KNO₂) =n(N₂) = 0,1 моль
V(N₂) = n*V(M) =  0,1 моль* 22,4 моль/л = 2,24 л.
n₂(NH₄Br) = 0,3306 моль - 0,1 моль = 0,2306 моль
m₂(NH₄Br) = 0,2306 моль * 98 г/моль = 22,6 г. -- бромид аммония, который не прореагировал с нитритом калия
ω(NH₄Br) = m(NH₄Br) : m(p-pa) = 22,6г/m(p-pa)*100%
m(p-pa) = 270 г + m(KNO₂) - m(N₂) 
m(N₂)= n*M = 0,1 моль* 28 г/моль = 2,8 г.
m(p-pa) = 270 г + m(KNO₂) - m(N₂) = 270 г. + 8,5 г. - 2,8 г. = 275,7 г.
ω(NH₄Br)  = 22,6г/275,7г. * 100% = 8,197% ≈ 8,2%

Answer 2

Calculation of Nitrogen Gas Volume and Ammonium Bromide Mass Fraction

To calculate the volume of nitrogen gas released and the mass fraction of ammonium bromide in the resulting solution, we need to use the given information.

Given: - Mass of potassium nitrite (KNO2) = 8.5 g - Mass of ammonium bromide (NH4Br) solution = 270 g - Mass fraction of ammonium bromide in the solution = 12%

To find the volume of nitrogen gas released, we need to determine the stoichiometry of the reaction between potassium nitrite and ammonium bromide. The balanced equation for this reaction is:

2 KNO2 + NH4Br → 2 KBr + N2 + 2 H2O

From the balanced equation, we can see that 2 moles of potassium nitrite react with 1 mole of ammonium bromide to produce 1 mole of nitrogen gas.

To calculate the moles of potassium nitrite, we can use its molar mass. The molar mass of potassium nitrite (KNO2) is approximately 85.1 g/mol.

Moles of potassium nitrite = Mass of potassium nitrite / Molar mass of potassium nitrite Moles of potassium nitrite = 8.5 g / 85.1 g/mol

To calculate the moles of nitrogen gas released, we can use the stoichiometry of the reaction.

Moles of nitrogen gas = Moles of potassium nitrite / 2

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

PV = nRT

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

Volume of nitrogen gas = Moles of nitrogen gas * (R * T) / P

To calculate the mass fraction of ammonium bromide in the resulting solution, we can use the formula:

Mass fraction = (Mass of ammonium bromide / Mass of solution) * 100

Now let's calculate the volume of nitrogen gas and the mass fraction of ammonium bromide using the given information.

Calculation:

1. Moles of potassium nitrite: - Molar mass of potassium nitrite (KNO2) = 85.1 g/mol - Moles of potassium nitrite = 8.5 g / 85.1 g/mol

2. Moles of nitrogen gas: - Moles of nitrogen gas = Moles of potassium nitrite / 2

3. Volume of nitrogen gas at STP: - Pressure (P) = 1 atm - Temperature (T) = 273.15 K - Ideal gas constant (R) = 0.0821 L·atm/(mol·K) - Volume of nitrogen gas = Moles of nitrogen gas * (R * T) / P

4. Mass fraction of ammonium bromide: - Mass fraction = (Mass of ammonium bromide / Mass of solution) * 100

Let's calculate the values step by step.

Calculation Results:

1. Moles of potassium nitrite: - Molar mass of potassium nitrite (KNO2) = 85.1 g/mol - Moles of potassium nitrite = 8.5 g / 85.1 g/mol = 0.1 mol 2. Moles of nitrogen gas: - Moles of nitrogen gas = Moles of potassium nitrite / 2 = 0.1 mol / 2 = 0.05 mol

3. Volume of nitrogen gas at STP: - Pressure (P) = 1 atm - Temperature (T) = 273.15 K - Ideal gas constant (R) = 0.0821 L·atm/(mol·K) - Volume of nitrogen gas = Moles of nitrogen gas * (R * T) / P - Volume of nitrogen gas = 0.05 mol * (0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm = 1.19 L

4. Mass fraction of ammonium bromide: - Mass fraction = (Mass of ammonium bromide / Mass of solution) * 100 - Mass fraction = (270 g * 0.12) / 270 g * 100 = 12%

Answer:

The volume of nitrogen gas released during the reaction is 1.19 L at standard temperature and pressure (STP). The mass fraction of ammonium bromide in the resulting solution is 12%.