Question

Какая масса кислорода потребуется для получения 0,5 моль оксида алюминия. (4Al + 3O2→2Al2O3)​

Answer 1

Ответ:

3,8....................

Answer 2

Calculating Oxygen Mass Required for Aluminum Oxide Production

To calculate the mass of oxygen required for the production of 0.5 moles of aluminum oxide (Al2O3) using the balanced chemical equation 4Al + 3O2 → 2Al2O3, we can use stoichiometry.

The molar ratio between aluminum and oxygen in the balanced chemical equation is 4 moles of aluminum to 3 moles of oxygen.

Therefore, for 4 moles of aluminum, 3 moles of oxygen are required. This means that for 1 mole of aluminum, 3/4 moles of oxygen are required.

Given that we need to find the mass of oxygen required for 0.5 moles of aluminum oxide, we can use the molar mass of oxygen to convert moles to grams.

Calculation

1. Molar Mass of Oxygen (O2): - The molar mass of oxygen (O2) is approximately 32 g/mol.

2. Moles of Oxygen Required: - For 1 mole of aluminum, 3/4 moles of oxygen are required. - For 0.5 moles of aluminum oxide, the moles of oxygen required can be calculated as: 0.5 moles Al2O3 * (3/4) moles O2/mole Al = 0.375 moles O2.

3. Mass of Oxygen Required: - Using the molar mass of oxygen, we can calculate the mass of oxygen required: Mass = moles * molar mass = 0.375 moles * 32 g/mol = 12 grams of oxygen.

Therefore, the mass of oxygen required for the production of 0.5 moles of aluminum oxide is 12 grams.

I hope this helps! If you have any further questions, feel free to ask.