Question

В результате реакции разложения гидроксида хрома(III) образуется оксид хрома(III) и вода.Определите массы продуктов реакции,полученных при разложении 2,06 кг гидроксида хрома(III) РЕШИТЕ ПОЖАЛУЙСТА!!!

Answer 1

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Answer 2

Reaction and Products

The reaction given is the decomposition of chromium(III) hydroxide into chromium(III) oxide and water. To determine the masses of the reaction products obtained from the decomposition of 2.06 kg of chromium(III) hydroxide, we need to know the molar masses of chromium(III) hydroxide, chromium(III) oxide, and water.

The molar mass of chromium(III) hydroxide (Cr(OH)3) can be calculated by summing the atomic masses of its constituent elements: - Chromium (Cr) has an atomic mass of 51.996 g/mol. - Hydrogen (H) has an atomic mass of 1.008 g/mol. - Oxygen (O) has an atomic mass of 16.00 g/mol.

Therefore, the molar mass of chromium(III) hydroxide is: (1 * 51.996 g/mol) + (3 * (1.008 g/mol + 16.00 g/mol)) = 97.996 g/mol.

The molar mass of chromium(III) oxide (Cr2O3) can be calculated similarly: - Chromium (Cr) has an atomic mass of 51.996 g/mol. - Oxygen (O) has an atomic mass of 16.00 g/mol.

Therefore, the molar mass of chromium(III) oxide is: (2 * 51.996 g/mol) + (3 * 16.00 g/mol) = 151.996 g/mol.

The molar mass of water (H2O) is: (2 * 1.008 g/mol) + 16.00 g/mol = 18.016 g/mol.

Calculation of Masses

To calculate the masses of the products obtained from the decomposition of 2.06 kg of chromium(III) hydroxide, we need to use stoichiometry and the molar masses of the reactants and products.

The balanced chemical equation for the reaction is: 2 Cr(OH)3 -> Cr2O3 + 3 H2O.

From the balanced equation, we can see that 2 moles of chromium(III) hydroxide decompose to form 1 mole of chromium(III) oxide and 3 moles of water.

To calculate the moles of chromium(III) hydroxide, we divide the given mass by its molar mass: 2.06 kg = 2060 g. Moles of chromium(III) hydroxide = 2060 g / 97.996 g/mol = 20.92 mol.

Using the stoichiometry of the balanced equation, we can determine the moles of chromium(III) oxide and water produced: - Moles of chromium(III) oxide = 20.92 mol / 2 = 10.46 mol. - Moles of water = 20.92 mol * 3 = 62.76 mol.

To calculate the masses of the products, we multiply the moles by their respective molar masses: - Mass of chromium(III) oxide = 10.46 mol * 151.996 g/mol = 1589.68 g. - Mass of water = 62.76 mol * 18.016 g/mol = 1129.89 g.

Therefore, the masses of the products obtained from the decomposition of 2.06 kg of chromium(III) hydroxide are approximately: - Mass of chromium(III) oxide: 1589.68 g. - Mass of water: 1129.89 g.

Please note that these values are approximate and may vary slightly due to rounding errors.