Question

Помогите пожалуйста! Даю 40 баллов! Для получения кислорода термическим разложением: а) перманганата калия КМnО4; б) оксида ртути НgО; в) хлората калия КСlO3 - взяли оди- наковые навески каждого вещества - 10 г. В результате какой из реакций выделится больший объем кирлорода? Напишите уравнения реакций и ре- шите задачу.

Answer 1

А)
  10 г                                   x л
2KMnO4=K2MnO4+MnO2+O2
2*158 г/моль                     22,4 л/моль

V(O2)=m(KMnO4)*Vm/M(KMnO4)=10 г*22,4 л/моль/2*158 г/моль=0,71 л.
Б)
  10 г           x л
2HgO=2Hg+O2
217 г/моль  22,4 л/моль
V(O2)=m(HgO)*Vm/M(HgO)=10 г*22,4 л/моль/217 г/моль=1,03 л.
В)
  10 г
2KClO3=2KCl+3O2
122,5 г/моль  3*22,4 л/моль
V(O2)=m(KClO3)*Vm/M(KClO3)=10 г*3*22,4 л/моль/122,5 г/моль=5,5 л.

Ответ: В.

Answer 2

Decomposition Reactions and Oxygen Production

To determine which reaction will produce a larger volume of oxygen, we need to examine the decomposition reactions of potassium permanganate (KMnO4), mercury oxide (HgO), and potassium chlorate (KClO3).

1. Decomposition of Potassium Permanganate (KMnO4): The decomposition of potassium permanganate can be represented by the following equation: ``` 2 KMnO4(s) -> K2MnO4(s) + MnO2(s) + O2(g) ``` According to this equation, for every 2 moles of KMnO4 decomposed, 1 mole of O2 gas is produced.

2. Decomposition of Mercury Oxide (HgO): The decomposition of mercury oxide can be represented by the following equation: ``` 2 HgO(s) -> 2 Hg(l) + O2(g) ``` According to this equation, for every 2 moles of HgO decomposed, 1 mole of O2 gas is produced.

3. Decomposition of Potassium Chlorate (KClO3): The decomposition of potassium chlorate can be represented by the following equation: ``` 2 KClO3(s) -> 2 KCl(s) + 3 O2(g) ``` According to this equation, for every 2 moles of KClO3 decomposed, 3 moles of O2 gas are produced.

Calculation of Oxygen Production

To determine which reaction will produce a larger volume of oxygen, we need to compare the number of moles of oxygen produced by each reaction. Since the mass of each substance is given as 10 grams, we can calculate the number of moles using their molar masses.

1. Potassium Permanganate (KMnO4): The molar mass of KMnO4 is 158.034 g/mol. Therefore, the number of moles of KMnO4 is: ``` moles of KMnO4 = mass of KMnO4 / molar mass of KMnO4 = 10 g / 158.034 g/mol ≈ 0.0633 mol ``` Since the stoichiometric coefficient of O2 in the equation is 1, the number of moles of O2 produced is also 0.0633 mol.

2. Mercury Oxide (HgO): The molar mass of HgO is 216.59 g/mol. Therefore, the number of moles of HgO is: ``` moles of HgO = mass of HgO / molar mass of HgO = 10 g / 216.59 g/mol ≈ 0.0461 mol ``` Since the stoichiometric coefficient of O2 in the equation is 1, the number of moles of O2 produced is also 0.0461 mol.

3. Potassium Chlorate (KClO3): The molar mass of KClO3 is 122.55 g/mol. Therefore, the number of moles of KClO3 is: ``` moles of KClO3 = mass of KClO3 / molar mass of KClO3 = 10 g / 122.55 g/mol ≈ 0.0816 mol ``` Since the stoichiometric coefficient of O2 in the equation is 3, the number of moles of O2 produced is 3 times the number of moles of KClO3: ``` moles of O2 = 3 * moles of KClO3 = 3 * 0.0816 mol ≈ 0.2448 mol ```

Conclusion

Comparing the number of moles of oxygen produced by each reaction, we find that the decomposition of potassium chlorate (KClO3) will result in a larger volume of oxygen. The reaction of potassium chlorate produces approximately 0.2448 moles of O2, while the reactions of potassium permanganate and mercury oxide produce only 0.0633 moles and 0.0461 moles of O2, respectively.

Therefore, the decomposition of potassium chlorate (KClO3) will yield the largest volume of oxygen among the three reactions.

Please note that the calculations provided are based on the given masses of the substances and the stoichiometry of the reactions.

Answer 3

Thermal Decomposition for Oxygen Production

To determine which substance will produce the largest volume of oxygen through thermal decomposition, we can examine the chemical equations for the reactions and calculate the molar volumes of oxygen gas produced by each substance.

Chemical Equations for Thermal Decomposition

1. Potassium Permanganate (KMnO4): - The thermal decomposition of potassium permanganate produces oxygen gas and manganese(IV) oxide. - The balanced chemical equation for this reaction is: ``` 2 KMnO4(s) → K2MnO4(s) + 3 O2(g) ```

2. Mercury(II) Oxide (HgO): - When mercury(II) oxide undergoes thermal decomposition, it yields oxygen gas and elemental mercury. - The balanced chemical equation for this reaction is: ``` 2 HgO(s) → 2 Hg(l) + O2(g) ```

3. Potassium Chlorate (KClO3): - The thermal decomposition of potassium chlorate results in the formation of oxygen gas and potassium chloride. - The balanced chemical equation for this reaction is: ``` 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ```

Molar Volumes of Oxygen Gas

To determine which substance will produce the largest volume of oxygen, we can calculate the molar volumes of oxygen gas produced by each substance using the ideal gas law, which states that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters.

- From the balanced chemical equations, we can see that for every 2 moles of potassium permanganate, 3 moles of oxygen gas are produced. - Similarly, for every 2 moles of potassium chlorate, 3 moles of oxygen gas are produced. - For every 2 moles of mercury(II) oxide, 1 mole of oxygen gas is produced.

Conclusion

Based on the molar volumes of oxygen gas produced by each substance, potassium permanganate (KMnO4) and potassium chlorate (KClO3) will produce the same volume of oxygen gas, which is greater than the volume produced by mercury(II) oxide (HgO).

Therefore, in the given scenario, the reactions of potassium permanganate and potassium chlorate will result in a larger volume of oxygen compared to the reaction of mercury(II) oxide.

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