Question

Рассчитайте массу соли и объем кислорода, которые образуются при разложении 123г перманганата калия.

Answer 1

Calculation of the Mass of Salt and Volume of Oxygen

To calculate the mass of salt and volume of oxygen formed during the decomposition of 123g of potassium permanganate (KMnO4), we need to consider the balanced chemical equation for the reaction.

The balanced chemical equation for the decomposition of potassium permanganate is:

2KMnO4 -> 2KOH + 2MnO2 + 3O2

From the balanced equation, we can see that for every 2 moles of KMnO4 decomposed, 3 moles of O2 are produced.

To calculate the mass of salt and volume of oxygen, we need to follow these steps:

Step 1: Calculate the number of moles of KMnO4: Given mass of KMnO4 = 123g Molar mass of KMnO4 = 158.034g/mol (from source Number of moles of KMnO4 = mass / molar mass = 123g / 158.034g/mol

Step 2: Calculate the number of moles of O2: From the balanced equation, we know that 2 moles of KMnO4 produce 3 moles of O2. So, the number of moles of O2 = (number of moles of KMnO4) * (3 moles of O2 / 2 moles of KMnO4)

Step 3: Calculate the mass of salt (KOH + MnO2): From the balanced equation, we know that 2 moles of KMnO4 produce 2 moles of KOH and 2 moles of MnO2. So, the mass of salt = (number of moles of KMnO4) * (molar mass of KOH + molar mass of MnO2)

Step 4: Calculate the volume of oxygen: To calculate the volume of oxygen, we need to use the ideal gas law equation: PV = nRT

Given: Pressure (P) = 1 atm (standard atmospheric pressure) Temperature (T) = 273.15 K (standard temperature) R = 0.0821 L.atm/mol.K (ideal gas constant)

Using the ideal gas law equation, we can calculate the volume (V) of oxygen: V = (number of moles of O2) * (R) * (T) / (P)

Now, let's calculate the mass of salt and volume of oxygen using the given information.

Step 1: Calculate the number of moles of KMnO4: Number of moles of KMnO4 = 123g / 158.034g/mol = 0.778 mol

Step 2: Calculate the number of moles of O2: Number of moles of O2 = (0.778 mol) * (3 mol O2 / 2 mol KMnO4) = 1.167 mol

Step 3: Calculate the mass of salt (KOH + MnO2): Molar mass of KOH = 56.1056 g/mol (from source Molar mass of MnO2 = 86.9368 g/mol (from source

Mass of salt = (0.778 mol) * (56.1056 g/mol + 86.9368 g/mol) = 107.9 g

Step 4: Calculate the volume of oxygen: V = (1.167 mol) * (0.0821 L.atm/mol.K) * (273.15 K) / (1 atm) = 27.3 L

Therefore, during the decomposition of 123g of potassium permanganate, approximately 107.9g of salt (KOH + MnO2) is formed, and the volume of oxygen produced is approximately 27.3 liters.