Question

При взаимодействии алюминия с кислородом образуется оксид алюминия. Рассчитайте количество вещества, массу и объём ( н.у. ) кислорода, который потребуется для взаимодействия с 27 г алюминия, содержащего 10% примесей?

Answer 1

Дано:

m( Al+ примесь) = 27 г

W( примесь)= 10%

Найти:

n(O2)-?

m(O2)-?

V(O2)-?

Решение:

4Al+3O2=2Al2O3

W(Al)= 100-10=90% или 0,9

m(Al) = W(Al)*m(Al+ примесь)= 0,9*27 г=24,3 г

n(Al)=m/M= 24,3 г:27 г/моль = 0,9 моль

M(Al) =27 г/моль

n(Al)=n(O2)=0,675 моль

m(O2)=n*M= 0,675 моль*32 г/моль=21,6 г

M(O2)= 16*2 =32 г/моль

V(O2)= n*Vm= 0,675 моль*22,4 л/моль15,12 л

Ответ: n(O2)= 0,675 моль; m(O2) =21,6 г; V(O2) =15,12 л

Answer 2

Calculation of the Amount of Oxygen Required for the Reaction

To calculate the amount of oxygen required for the reaction with 27 g of aluminum, we need to consider the stoichiometry of the reaction. The balanced chemical equation for the reaction between aluminum and oxygen is:

2 Al + 3 O2 → 2 Al2O3

From the balanced equation, we can see that 2 moles of aluminum react with 3 moles of oxygen to form 2 moles of aluminum oxide (Al2O3).

To find the amount of oxygen required, we can use the molar mass of aluminum and the given mass of aluminum.

The molar mass of aluminum (Al) is approximately 26.98 g/mol.

First, let's calculate the number of moles of aluminum:

Number of moles of aluminum = Mass of aluminum / Molar mass of aluminum Number of moles of aluminum = 27 g / 26.98 g/mol

Now, let's calculate the number of moles of oxygen required using the stoichiometry of the reaction:

Number of moles of oxygen = (Number of moles of aluminum / 2) * (3 moles of oxygen / 2 moles of aluminum) Number of moles of oxygen = (27 g / 26.98 g/mol) * (3 mol O2 / 2 mol Al)

Calculation of the Mass and Volume of Oxygen

To calculate the mass and volume of oxygen, we need to use the ideal gas law, which states that the volume of a gas is directly proportional to the number of moles of the gas.

The molar volume of a gas at standard temperature and pressure (STP) is 22.4 L/mol.

Let's calculate the mass of oxygen:

Mass of oxygen = Number of moles of oxygen * Molar mass of oxygen Mass of oxygen = (27 g / 26.98 g/mol) * (3 mol O2 / 2 mol Al) * Molar mass of oxygen

To calculate the volume of oxygen, we can use the molar volume at STP:

Volume of oxygen = Number of moles of oxygen * Molar volume at STP Volume of oxygen = (27 g / 26.98 g/mol) * (3 mol O2 / 2 mol Al) * 22.4 L/mol

Now, let's substitute the values and calculate the mass and volume of oxygen required.

Note: The molar mass of oxygen is approximately 32 g/mol.

Calculation:

Number of moles of aluminum = 27 g / 26.98 g/mol = 1.001 mol Number of moles of oxygen = (1.001 mol) * (3 mol O2 / 2 mol Al) = 1.502 mol

Mass of oxygen = (1.502 mol) * (32 g/mol) = 48.064 g Volume of oxygen = (1.502 mol) * (22.4 L/mol) = 33.6528 L

Therefore, the amount of substance (moles) of oxygen required is approximately 1.502 mol, the mass of oxygen required is approximately 48.064 g, and the volume of oxygen required at STP is approximately 33.6528 L.

Please note that these calculations assume ideal conditions and may not account for all factors that could affect the reaction.