Question

закрытом сосуде смешаны оксид углерода (II) массой 30 г и кислород массой 20 г. Вычислите массу получившегося углекислого газа. Какой газ останется в избытке? Какова масса этого газа?​

Answer 1

n(O2)=20:32=0,625(недостаток)

n(CO)=30:28≈1моль(избыток)

m(CO2)=30+20=50г

Ответ: m(CO2)=50г; m(CO)=30г(избыток)

Answer 2

Calculation of the Mass of Carbon Dioxide Produced

To calculate the mass of carbon dioxide produced when carbon monoxide (CO) and oxygen (O2) react, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.

Given: - Mass of carbon monoxide (CO) = 30 g - Mass of oxygen (O2) = 20 g

To find the limiting reactant, we need to compare the number of moles of each reactant. We can use the molar mass of each substance to convert the mass to moles.

The molar mass of carbon monoxide (CO) is 28 g/mol, and the molar mass of oxygen (O2) is 32 g/mol.

The number of moles of carbon monoxide (CO) can be calculated as follows: Number of moles of CO = Mass of CO / Molar mass of CO = 30 g / 28 g/mol = 1.071 mol (approximately)

The number of moles of oxygen (O2) can be calculated as follows: Number of moles of O2 = Mass of O2 / Molar mass of O2 = 20 g / 32 g/mol = 0.625 mol (approximately)

Since the reaction between carbon monoxide (CO) and oxygen (O2) occurs in a 1:1 ratio, the number of moles of carbon dioxide (CO2) produced will be equal to the number of moles of the limiting reactant.

In this case, oxygen (O2) is the limiting reactant because it has fewer moles compared to carbon monoxide (CO).

Therefore, the number of moles of carbon dioxide (CO2) produced is 0.625 mol.

To calculate the mass of carbon dioxide (CO2) produced, we can use the molar mass of carbon dioxide, which is 44 g/mol.

Mass of carbon dioxide (CO2) = Number of moles of CO2 * Molar mass of CO2 = 0.625 mol * 44 g/mol = 27.5 g (approximately)

Therefore, the mass of the carbon dioxide produced is approximately 27.5 grams.

Excess Gas and Its Mass

To determine which gas is in excess, we need to compare the stoichiometric ratio of the reactants. The balanced chemical equation for the reaction between carbon monoxide (CO) and oxygen (O2) to form carbon dioxide (CO2) is:

2 CO + O2 -> 2 CO2

From the balanced equation, we can see that for every 2 moles of carbon monoxide (CO), we need 1 mole of oxygen (O2) to react completely.

In this case, we have 1.071 moles of carbon monoxide (CO) and 0.625 moles of oxygen (O2).

Since the stoichiometric ratio is 2:1, we can calculate the number of moles of oxygen (O2) required to react with the given moles of carbon monoxide (CO) as follows:

Number of moles of O2 required = (Number of moles of CO * 1) / 2 = (1.071 mol * 1) / 2 = 0.536 mol (approximately)

Comparing this with the actual number of moles of oxygen (O2) given (0.625 mol), we can conclude that oxygen (O2) is in excess.

The gas that remains in excess is oxygen (O2).

To calculate the mass of the excess gas (oxygen), we can use its molar mass, which is 32 g/mol.

Mass of excess oxygen (O2) = Number of moles of excess O2 * Molar mass of O2 = (0.625 mol - 0.536 mol) * 32 g/mol = 2.848 g (approximately)

Therefore, the mass of the excess oxygen gas is approximately 2.848 grams.

Please note that the calculations provided are approximate and may vary slightly depending on the actual values used.