Question

Определите объем кислорода выделившегося при термическом разложении 51,2 г перманганата калия содержащего 5% примесей

Answer 1

2KMnO4 → K2MnO4 + MnO2 + O2↑
1) W%(KMnO4 без примесе й) = 100% - 5% = 95%
2) m(KMnO4 без примесей) = 95% * 51,2г : 100% = 48,64г
3) n(KMnO4) = 48,64г : 158г/моль ≈ 0,31моль
4) n(O2) = 0,31моль : 2 = 0,155моль
5) V(O2) = 0,155моль * 22,4л/моль = 3,472л ≈ 3,5л

Answer 2

Calculation of Oxygen Volume Released

To determine the volume of oxygen released during the thermal decomposition of potassium permanganate, we need to consider the stoichiometry of the reaction. The balanced equation for the thermal decomposition of potassium permanganate (KMnO4) is as follows:

2 KMnO4(s) -> 2 K2MnO4(s) + O2(g)

From the equation, we can see that for every 2 moles of KMnO4, 1 mole of O2 is released. To calculate the volume of oxygen, we need to convert the mass of potassium permanganate to moles and then use the mole ratio to determine the number of moles of oxygen.

Given: - Mass of potassium permanganate (KMnO4) = 51.2 g - Percentage of impurities = 5%

First, we need to calculate the mass of pure KMnO4 by subtracting the mass of impurities from the total mass:

Mass of pure KMnO4 = Total mass of KMnO4 - Mass of impurities = 51.2 g - (5% of 51.2 g) = 51.2 g - (0.05 * 51.2 g) = 51.2 g - 2.56 g = 48.64 g

Next, we calculate the number of moles of KMnO4 using its molar mass:

Molar mass of KMnO4 = 39.1 g/mol (K) + 54.9 g/mol (Mn) + 16.0 g/mol (O) * 4 = 39.1 g/mol + 54.9 g/mol + 64.0 g/mol = 158.0 g/mol

Number of moles of KMnO4 = Mass of KMnO4 / Molar mass of KMnO4 = 48.64 g / 158.0 g/mol ≈ 0.308 moles

Since the mole ratio between KMnO4 and O2 is 2:1, the number of moles of O2 released is half the number of moles of KMnO4:

Number of moles of O2 = 0.308 moles / 2 = 0.154 moles

To calculate the volume of oxygen, we can use the ideal gas law:

PV = nRT

Where: P = pressure (assume standard pressure, 1 atm) V = volume n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (assume room temperature, around 298 K)

Rearranging the equation, we can solve for V:

V = (nRT) / P

Substituting the values:

V = (0.154 moles * 0.0821 L·atm/(mol·K) * 298 K) / 1 atm ≈ 3.82 L

Therefore, the volume of oxygen released during the thermal decomposition of 51.2 g of potassium permanganate, including 5% impurities, is approximately 3.82 liters.

Please note that the above calculation assumes ideal conditions and may not account for any deviations from the ideal gas law or other factors that may affect the actual volume of oxygen released.