Question

1.Термохимичекое уравнение сгорания аммиака имеет вид:4NH3+3O2=2N2+6H2O+1528кДжПри образовании 11.2 л （н. у. ）азота1.Поглащается 382 кДж теплоты2.Выделяется 382 кДж теплоты 3.Поглащается 764 кДж теплоты4.Выделяется 764 кДж теплоты 2.Согласно термохимическому уравнению реакции3Cu+8HNO3=3Cu(NO3)2+2NO+4H2O+358 кДжпри получении 15.68 л （н. у. ）оксида азота（2）кол-во выделившейся теплоты будет равно:1.358 кДж2.716 кДж3.125.3 кДж4.22.4 кДжМожно еще пошаговое обьясненение решения.

Answer 1

2.
Mr (2NO)=44,8л/моль
15,68л-х
44,8л/моль-358
x=15,68×358/44,8=125,3 кДж

Answer 2

Thermochemical Equations for Combustion of Ammonia and Nitric Acid

The thermochemical equation for the combustion of ammonia (NH3) is:

4NH3 + 3O2 → 2N2 + 6H2O + 1528 kJ [[1]]

This equation shows that when 4 moles of ammonia react with 3 moles of oxygen, it produces 2 moles of nitrogen gas, 6 moles of water, and releases 1528 kJ of heat energy.

The given information states that 11.2 liters of nitrogen gas are formed at standard conditions (n.u.). We need to determine the amount of heat energy absorbed or released during this reaction.

To solve this, we can use the concept of stoichiometry and the given thermochemical equation. The coefficients in the balanced equation represent the mole ratios between the reactants and products.

Since the equation states that 4 moles of ammonia produce 2 moles of nitrogen gas, we can set up a proportion:

4 moles NH3 / 2 moles N2 = x moles NH3 / 11.2 liters N2

Simplifying the proportion, we find:

x = (4/2) * (11.2 liters N2) = 22.4 liters NH3

Therefore, 22.4 liters of ammonia are required to produce 11.2 liters of nitrogen gas.

Now, let's determine the amount of heat energy released during the reaction.

The given options are:

1. 358 kJ 2. 716 kJ 3. 125.3 kJ 4. 22.4 kJ

To find the correct answer, we need to calculate the heat of reaction using the given thermochemical equation for the combustion of ammonia.

The equation states that the combustion of 4 moles of ammonia releases 1528 kJ of heat energy. Therefore, the heat of reaction for the combustion of 1 mole of ammonia is:

1528 kJ / 4 moles NH3 = 382 kJ/mole NH3

Since we have determined that 22.4 liters of ammonia are required to produce 11.2 liters of nitrogen gas, we can calculate the heat of reaction for this amount of ammonia:

382 kJ/mole NH3 * 22.4 moles NH3 = 8564.8 kJ

Therefore, the correct answer is:

4. 22.4 kJ

The amount of heat energy released during the reaction to produce 11.2 liters of nitrogen gas is 22.4 kJ.

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