Question

Расчитайте кол-во вещества и массу кислорода, образуещегося при разложении 24.5 г хлорида калия KClO3

Answer 1

Реакция разложения:
2KClO3 = 2KCl+3O2

Масса 2KClO3 - 24,5г.
Молекулярная масса (Mr) = 245 г/моль (с учетом коэффициентов)
Масса O2 - х грамм
Mr(O2) = 96 г/моль
Составляем пропорцию:
24,5/245 = х/96
Отсюда находим х: х=(24,5*96)/245 = 9,6 (грамм).
Мы нашли массу кислорода, теперь находим кол.-во в-ва.
v(O2)=m/Mr=m/Mr=9,6/96=0,1 (моль).

Answer 2

24,5:5=4.9(кол-во вещества
4.9*3(по правилу)=14.7(масса кислорода в граммах)

Answer 3

Calculating the Amount of Oxygen Formed from the Decomposition of Potassium Chlorate (KClO3)

To calculate the amount of substance and the mass of oxygen formed during the decomposition of 24.5 g of potassium chlorate (KClO3), we can use stoichiometry and the molar mass of oxygen.

Step 1: Write the Balanced Chemical Equation The decomposition of potassium chlorate can be represented by the following balanced chemical equation: ``` 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ```

Step 2: Calculate the Amount of Oxygen From the balanced chemical equation, we can see that 2 moles of potassium chlorate produce 3 moles of oxygen. Using the given mass of potassium chlorate (24.5 g), we can calculate the amount of oxygen produced.

First, we need to find the molar mass of potassium chlorate (KClO3): - Molar mass of K = 39.10 g/mol - Molar mass of Cl = 35.45 g/mol - Molar mass of O = 16.00 g/mol - Molar mass of KClO3 = 39.10 + 35.45 + 3(16.00) = 122.55 g/mol

Now, we can calculate the moles of KClO3: ``` moles of KClO3 = mass / molar mass moles of KClO3 = 24.5 g / 122.55 g/mol moles of KClO3 ≈ 0.20 mol ```

Using the stoichiometry of the balanced chemical equation, we can calculate the moles of oxygen produced: ``` moles of O2 = (moles of KClO3) × (3 moles of O2 / 2 moles of KClO3) moles of O2 = 0.20 mol × (3 / 2) moles of O2 = 0.30 mol ```

Step 3: Calculate the Mass of Oxygen Finally, we can calculate the mass of oxygen produced: ``` mass of O2 = moles of O2 × molar mass of O2 mass of O2 = 0.30 mol × 32.00 g/mol mass of O2 = 9.60 g ```

Conclusion: The amount of substance of oxygen formed during the decomposition of 24.5 g of potassium chlorate is approximately 0.30 moles, and the mass of oxygen formed is approximately 9.60 grams.