Question

Сколько выделится теплоты при сгорании 17 кг аммиака Сколько литров кислорода (н. у.) Вступает в реакци, если выделяется 764.52 кДж теплоты Сколько литров азота (н. у.) Образуется , если выделятся 4587,12 кДж теплоты

Answer 1

Много рта к воры овтру ВОВ провожала Шаталов автостопа латала латала в

Answer 2

Calculating Heat Release from Ammonia Combustion

To calculate the heat released from the combustion of 17 kg of ammonia, we can use the balanced chemical equation for the combustion of ammonia:

2NH3 + 3O2 → 2NO + 3H2O

The molar mass of ammonia (NH3) is approximately 17.031 g/mol, and the molar mass of oxygen (O2) is approximately 32 g/mol.

First, let's calculate the moles of ammonia and oxygen involved in the reaction:

Moles of NH3 = Mass of NH3 / Molar mass of NH3 Moles of O2 = Moles of NH3 * (3/2)

Then, we can use the enthalpy change of the reaction to calculate the heat released:

ΔH = 2 * 46.11 kJ/mol (for the combustion of NH3)

Finally, we can use the moles of ammonia and oxygen to calculate the heat released:

Heat released = Moles of NH3 * ΔH

Let's calculate the moles of ammonia and oxygen and then the heat released.

Calculating Moles of Ammonia and Oxygen

Moles of NH3: Moles of NH3 = 17000 g / 17.031 g/mol Moles of NH3 = 998.53 mol

Moles of O2: Moles of O2 = 998.53 mol * (3/2) Moles of O2 = 1497.79 mol

Calculating Heat Released

Heat released: Heat released = 998.53 mol * 2 * 46.11 kJ/mol Heat released = 919,999.77 kJ

So, the heat released from the combustion of 17 kg of ammonia is approximately 919,999.77 kJ.

Calculating Volume of Oxygen and Nitrogen Involved

To calculate the volume of oxygen and nitrogen involved, we can use the ideal gas law:

PV = nRT

Where: P = pressure (at standard conditions, 1 atm) V = volume n = moles R = ideal gas constant (0.0821 L·atm/mol·K) T = temperature (at standard conditions, 273 K)

We can rearrange the equation to solve for volume:

V = (nRT) / P

Volume of Oxygen (NTP)

Given: Heat released = 764.52 kJ Moles of O2 = 1497.79 mol P = 1 atm T = 273 K R = 0.0821 L·atm/mol·K

Volume of oxygen: V(O2) = (1497.79 mol * 0.0821 L·atm/mol·K * 273 K) / 1 atm V(O2) ≈ 33,073.67 L

So, the volume of oxygen (at standard conditions) involved in the reaction is approximately 33,073.67 liters.

Volume of Nitrogen (NTP)

Given: Heat released = 4587.12 kJ Moles of N2 = (1497.79 mol * 2) / 3 P = 1 atm T = 273 K R = 0.0821 L·atm/mol·K

Volume of nitrogen: V(N2) = ((1497.79 mol * 2) / 3 * 0.0821 L·atm/mol·K * 273 K) / 1 atm V(N2) ≈ 22,048.20 L

So, the volume of nitrogen (at standard conditions) formed in the reaction is approximately 22,048.20 liters.