Question

Сколько литров кислорода необходимо для получения 36 грамм оксида алюминия. АL + О2 → АL2О3

Answer 1

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Answer 2

Calculating the Amount of Oxygen Needed to Produce 36 Grams of Aluminum Oxide (Al2O3)

To calculate the amount of oxygen needed to produce 36 grams of aluminum oxide (Al2O3) from aluminum (Al), we can use the balanced chemical equation for the reaction:

2 Al + 3 O2 → 2 Al2O3

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

The molar mass of aluminum oxide (Al2O3) is 101.960 g/mol Therefore, 36 grams of aluminum oxide is equivalent to:

36 g / 101.960 g/mol = 0.352 moles of Al2O3

According to the balanced chemical equation, 2 moles of aluminum oxide are produced from 3 moles of oxygen. Therefore, the moles of oxygen needed can be calculated as:

0.352 moles of Al2O3 * (3 moles of O2 / 2 moles of Al2O3) = 0.528 moles of O2

Now, to convert moles of oxygen to liters, we can use the ideal gas law:

PV = nRT

Where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant - T = temperature

Given that the temperature and pressure are not specified, we can assume standard temperature and pressure (STP) for simplicity, which are 0°C (273.15 K) and 1 atm, respectively.

At STP, 1 mole of any gas occupies 22.4 liters. Therefore, the amount of oxygen needed can be calculated as:

0.528 moles of O2 * 22.4 L/mol = 11.83 liters of O2

So, to produce 36 grams of aluminum oxide (Al2O3) from aluminum (Al), approximately 11.83 liters of oxygen are needed.

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