Question

Какой объем кислорода необходим для окисления 5,4 грамм алюминия. Какое количество вещества продукта образуется?

Answer 1

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Answer 2

Calculating the Volume of Oxygen Needed for the Oxidation of Aluminum

To calculate the volume of oxygen needed for the oxidation of 5.4 grams of aluminum, we can use the balanced chemical equation for the reaction between aluminum and oxygen.

The balanced chemical equation for the reaction is: 4Al + 3O2 → 2Al2O3

From the equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.

Using the molar mass of aluminum (Al) and oxygen (O2), we can calculate the volume of oxygen needed for the reaction.

Step 1: Calculate the Moles of Aluminum

The molar mass of aluminum (Al) is approximately 27 g/mol.

The number of moles of aluminum can be calculated using the formula: moles = mass / molar mass

Substituting the given mass of aluminum (5.4 grams) into the formula: moles of Al = 5.4 g / 27 g/mol = 0.2 moles

Step 2: Calculate the Moles of Oxygen

From the balanced chemical equation, we know that 4 moles of aluminum react with 3 moles of oxygen.

Using the mole ratio, we can calculate the moles of oxygen needed: moles of O2 = (moles of Al) * (3 moles of O2 / 4 moles of Al) moles of O2 = 0.2 moles * (3/4) = 0.15 moles

Step 3: Calculate the Volume of Oxygen

The volume of oxygen needed can be calculated using the ideal gas law: V = nRT/P

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

Assuming standard temperature and pressure (STP), where T = 273 K and P = 1 atm, we can calculate the volume of oxygen needed.

V = (0.15 moles) * (0.0821 L·atm/mol·K) * (273 K) / (1 atm) V ≈ 3.7 liters

Therefore, approximately 3.7 liters of oxygen are needed for the oxidation of 5.4 grams of aluminum.

Calculating the Amount of Product Formed

The balanced chemical equation for the reaction between aluminum and oxygen is: 4Al + 3O2 → 2Al2O3

From the balanced equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.

Using the mole ratio, we can calculate the moles of aluminum oxide produced: moles of Al2O3 = (moles of O2) * (2 moles of Al2O3 / 3 moles of O2) moles of Al2O3 = 0.15 moles * (2/3) = 0.1 moles

Therefore, 0.1 moles of aluminum oxide are produced from the reaction.

In summary, approximately 3.7 liters of oxygen are needed for the oxidation of 5.4 grams of aluminum, and 0.1 moles of aluminum oxide are produced as a result of the reaction.