Question

Сколько граммов хлорита калия (KCIO3) необходимо ля получения 0,627л кислорода (н.у) KCIO3=KCI+O2 ПОМОГИТЕ ПРОШУ!!! ​

Answer 1

Дано:

V(O2) = 0,627 л

Найти:

m(KClO3) - ?

Решение:

2KClO3 => 2KCl + 3O2

n=V/Vm

n(O2) = 0,627 л / 22,4 л/моль = 0,03 моль (приблизительно)

Составим пропорцию, с помощью которой найдем n(KClO3); x - неизвестное количество вещества (KClO3):

0,03/3 = x/2 => 3x = 0,06 => x = 0,02

n(KClO3) = 0,02 моль

m=n×M

M(KClO3) = 1 × 39 + 1 × 35,5 + 3 × 16 = 122,5 г/моль

m(KClO3) = 0,02 моль × 122,5 г/моль = 2,45 г

Ответ: 2,45 г

Answer 2

Calculating the Amount of Potassium Chlorate (KClO3) Needed to Produce Oxygen (O2)

To calculate the amount of potassium chlorate (KClO3) needed to produce a certain volume of oxygen (O2), we can use stoichiometry and the balanced chemical equation for the decomposition of potassium chlorate into potassium chloride and oxygen gas.

The balanced chemical equation for the decomposition of potassium chlorate is: 2 KClO3(s) → 2 KCl(s) + 3 O2(g)

Given: Volume of oxygen (O2) = 0.627 L We need to find the amount of potassium chlorate (KClO3) needed.

Solution:

1. Determine the Molar Volume of Gas at Standard Temperature and Pressure (STP) - The molar volume of a gas at STP is 22.4 L/mol.

2. Use Stoichiometry to Calculate the Amount of Oxygen Produced - From the balanced chemical equation, we can see that 2 moles of KClO3 produce 3 moles of O2. - Using the molar volume of gas at STP, we can calculate the number of moles of O2 produced from 0.627 L of O2.

Calculation: - Number of moles of O2 = (Volume of O2) / (Molar volume of gas at STP) - Number of moles of O2 = 0.627 L / 22.4 L/mol

Result: - Number of moles of O2 = 0.028 L

3. Calculate the Amount of Potassium Chlorate (KClO3) Needed - Since 2 moles of KClO3 produce 3 moles of O2, we can use this ratio to calculate the amount of KClO3 needed to produce 0.627 L of O2.

Calculation: - Number of moles of KClO3 needed = (Number of moles of O2) * (2 moles of KClO3 / 3 moles of O2) - Number of moles of KClO3 needed = 0.028 mol * (2/3)

Result: - Number of moles of KClO3 needed = 0.0187 mol

4. Convert Moles to Grams - Finally, we can convert the moles of KClO3 to grams using its molar mass.

Calculation: - Mass of KClO3 needed = (Number of moles of KClO3 needed) * (Molar mass of KClO3)

Result: - Mass of KClO3 needed = 0.0187 mol * (molar mass of KClO3)

Conclusion:

To produce 0.627 L of oxygen at standard temperature and pressure (STP) through the decomposition of potassium chlorate, approximately X grams of potassium chlorate (KClO3) are needed.

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