Question

Дана схема реакции (без коэффициентов): NH3 (г) + O2 (г) -> NO (г) H2OВ реакцию вступает 32 литра аммиака с достаточным количеством кислорода. Сколько литров оксида азота(II) при этом можно получить? Объёмы газообразных веществ измерены при нормальных условиях.1. 962. 643. 324. 16

Answer 1

Ответ:

32                лрпап                        

Объяснение:

Answer 2

Reaction Scheme and Given Information

The given reaction scheme is as follows: NH3 (g) + O2 (g) -> NO (g) + H2O

It is stated that 32 liters of ammonia (NH3) react with an excess amount of oxygen (O2). We need to determine the volume of nitrogen oxide (NO) that can be produced in this reaction. The volumes of the gaseous substances are measured under normal conditions.

Solution

To solve this problem, we need to use the concept of stoichiometry, which relates the ratios of reactants and products in a chemical reaction.

From the balanced equation, we can see that the stoichiometric ratio between NH3 and NO is 1:1. This means that for every 1 mole of NH3, we will produce 1 mole of NO.

To find the volume of NO, we need to convert the given volume of NH3 to moles using the ideal gas law. The ideal gas law equation is given by:

PV = nRT

Where: P = pressure (in atm) V = volume (in liters) n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (in Kelvin)

Since the volumes are measured under normal conditions, we can assume that the pressure is 1 atm and the temperature is 273 K.

Using the ideal gas law, we can calculate the number of moles of NH3:

n(NH3) = (PV)/(RT) = (1 atm * 32 L) / (0.0821 L·atm/(mol·K) * 273 K) = 1.23 moles

Since the stoichiometric ratio between NH3 and NO is 1:1, the number of moles of NO produced will also be 1.23 moles.

To convert the moles of NO to volume, we can use the ideal gas law again:

V(NO) = (n(NO) * RT) / P = (1.23 moles * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm = 27.5 liters

Therefore, the volume of nitrogen oxide (NO) that can be produced in this reaction is 27.5 liters.

Answer

The volume of nitrogen oxide (NO) that can be produced in this reaction is 27.5 liters.