Question

какая масса алюминия потребуется для получения железа методом алюмотермии из 20г оксида железа (3) содержащего 10 процентов примисей

Answer 1

Ответ:

Объяснение:

50 кг

Answer 2

Calculation of Aluminum Required for Iron Production by Aluminothermic Method

To calculate the mass of aluminum required for the production of iron using the aluminothermic method, we need to consider the stoichiometry of the reaction and the molar masses of the compounds involved.

The balanced chemical equation for the reaction between aluminum and iron(III) oxide is as follows:

2Al + Fe2O3 -> 2Fe + Al2O3

From the equation, we can see that 2 moles of aluminum react with 1 mole of iron(III) oxide to produce 2 moles of iron and 1 mole of aluminum oxide.

To determine the mass of aluminum required, we need to convert the given mass of iron(III) oxide to moles and then use the stoichiometric ratio to calculate the mass of aluminum.

Given: - Mass of iron(III) oxide (Fe2O3) = 20 g - Iron(III) oxide contains 10% impurities

First, we need to calculate the mass of pure iron(III) oxide by subtracting the mass of impurities:

Mass of pure iron(III) oxide = Mass of iron(III) oxide - Mass of impurities = 20 g - (10% of 20 g) = 20 g - 2 g = 18 g

Next, we calculate the number of moles of iron(III) oxide:

Number of moles of iron(III) oxide = Mass of iron(III) oxide / Molar mass of iron(III) oxide

The molar mass of iron(III) oxide (Fe2O3) can be calculated by summing the molar masses of iron (Fe) and oxygen (O):

Molar mass of Fe2O3 = (2 * Molar mass of Fe) + (3 * Molar mass of O)

Using the atomic masses of iron (Fe = 55.845 g/mol) and oxygen (O = 16.00 g/mol):

Molar mass of Fe2O3 = (2 * 55.845 g/mol) + (3 * 16.00 g/mol) = 111.69 g/mol + 48.00 g/mol = 159.69 g/mol

Now we can calculate the number of moles of iron(III) oxide:

Number of moles of iron(III) oxide = 18 g / 159.69 g/mol

Finally, we use the stoichiometric ratio from the balanced equation to calculate the mass of aluminum required:

Mass of aluminum = (2 moles of aluminum / 1 mole of iron(III) oxide) * (Molar mass of aluminum) * (Number of moles of iron(III) oxide)

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

Let's calculate the mass of aluminum required:

Mass of aluminum = (2 / 1) * (26.98 g/mol) * (18 g / 159.69 g/mol)

Mass of aluminum required = 6.04 g

Therefore, approximately 6.04 grams of aluminum would be required to produce iron using the aluminothermic method from 20 grams of iron(III) oxide containing 10% impurities.

Please note that the calculation assumes ideal conditions and does not account for any losses or inefficiencies in the reaction process.