Question

В закрытом сосуде смешали 30 г оксида азота (2) и 20 г кислорода. вычислите массу продукта реакции. какой газ останется в избытке?

Answer 1

2NO + O2 = 2NO2
1. Находим количества веществ, взятых для реакции, зная их массу и молярную массу:
n(NO) = m / M = 30 / 30 = 1 моль
n(O2) = m / M = 20 / 32 = 0.625 моль
2. Составив уравнение реакции, мы видим:
n(NO) / n(O2) = 2 / 1, т.е. n(O2) = n(NO) / 2;
n(O2) = 1 / 2 = 0.5 моль 
т.е. для реакции с 30 г оксида азота требуется 0,5 моль кислорода, а взято его - 0,625 моль. Следовательно, кислород взят в избытке. Расчет массы продукта реакции будем проводить с использованием количества вещества оксида азота.
3. По уравнению реакции видно, что n(NO) / n(NO2) = 2 / 2, (количества равны), значит:
n(NO2) = n(NO) = 1 моль
4. Зная количество NO2 и его молярную массу, находим его массу:
m(NO2) = n x M = 1 x 46 = 46 г

Answer 2

Calculation of the Mass of the Reaction Product

To calculate the mass of the reaction product, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

In this case, we have 30 g of nitrogen oxide (NO2) and 20 g of oxygen (O2). The balanced chemical equation for the reaction between nitrogen oxide and oxygen is:

2 NO2 + O2 → 2 NO3

According to the stoichiometry of the reaction, 2 moles of nitrogen oxide react with 1 mole of oxygen to produce 2 moles of nitrogen trioxide.

To find the limiting reactant, we can compare the number of moles of each reactant. The molar mass of nitrogen oxide (NO2) is approximately 46 g/mol, and the molar mass of oxygen (O2) is approximately 32 g/mol.

The number of moles of nitrogen oxide is calculated as follows:

Number of moles of nitrogen oxide = mass of nitrogen oxide / molar mass of nitrogen oxide

Number of moles of nitrogen oxide = 30 g / 46 g/mol ≈ 0.652 moles

The number of moles of oxygen is calculated as follows:

Number of moles of oxygen = mass of oxygen / molar mass of oxygen

Number of moles of oxygen = 20 g / 32 g/mol ≈ 0.625 moles

Since the stoichiometry of the reaction is 2:1, we can see that the number of moles of oxygen is less than half of the number of moles of nitrogen oxide. Therefore, oxygen is the limiting reactant.

To calculate the mass of the reaction product, we can use the number of moles of the limiting reactant. Since 1 mole of oxygen reacts to produce 2 moles of nitrogen trioxide, the number of moles of nitrogen trioxide formed is:

Number of moles of nitrogen trioxide = 2 * number of moles of oxygen

Number of moles of nitrogen trioxide = 2 * 0.625 moles = 1.25 moles

The molar mass of nitrogen trioxide (NO3) is approximately 62 g/mol. Therefore, the mass of the reaction product (nitrogen trioxide) is:

Mass of nitrogen trioxide = number of moles of nitrogen trioxide * molar mass of nitrogen trioxide

Mass of nitrogen trioxide = 1.25 moles * 62 g/mol ≈ 77.5 g

Therefore, the mass of the reaction product (nitrogen trioxide) is approximately 77.5 grams.

Excess Gas Remaining

To determine which gas remains in excess, we need to compare the stoichiometric ratio of the reactants. In this case, the stoichiometry of the reaction is 2:1, meaning that 2 moles of nitrogen oxide react with 1 mole of oxygen.

Since we have determined that oxygen is the limiting reactant, we can calculate the amount of nitrogen oxide that would be required to react completely with the available oxygen.

The number of moles of nitrogen oxide required to react with the available oxygen is:

Number of moles of nitrogen oxide required = 0.5 * number of moles of oxygen

Number of moles of nitrogen oxide required = 0.5 * 0.625 moles = 0.3125 moles

The initial amount of nitrogen oxide we have is 0.652 moles. Therefore, there is an excess of nitrogen oxide.

To calculate the mass of the excess nitrogen oxide, we can use the number of moles of the excess nitrogen oxide and its molar mass. The molar mass of nitrogen oxide (NO2) is approximately 46 g/mol.

Mass of excess nitrogen oxide = (number of moles of nitrogen oxide - number of moles of nitrogen oxide required) * molar mass of nitrogen oxide

Mass of excess nitrogen oxide = (0.652 moles - 0.3125 moles) * 46 g/mol ≈ 14.7 g

Therefore, the gas that remains in excess is nitrogen oxide, with a mass of approximately 14.7 grams.

In summary: - The mass of the reaction product (nitrogen trioxide) is approximately 77.5 grams. - The gas that remains in excess is nitrogen oxide, with a mass of approximately 14.7 grams.