Question

При действии 95 г водного раствора этанола на избыток натрия выделился га на сжигание которого требуется 25,2 л (н.у.) кислорода. Найдите массы этилена и воды,которые нужны для получения указанной массы раствора этанола.

Answer 1

Calculation of Masses of Ethylene and Water

To calculate the masses of ethylene and water required to obtain the specified mass of the ethanol solution, we need to use the balanced chemical equation for the reaction between ethanol and excess sodium:

C2H5OH + 2Na → C2H4 + 2NaOH

From the equation, we can see that 1 mole of ethanol (C2H5OH) reacts to produce 1 mole of ethylene (C2H4). Additionally, we know that the volume of oxygen (O2) required for the combustion of the produced gas is 25.2 liters at standard temperature and pressure (n.t.p).

To find the masses of ethylene and water, we need to follow these steps:

1. Calculate the number of moles of oxygen gas (O2) required for the combustion reaction using the ideal gas law. 2. Use the stoichiometry of the balanced equation to determine the number of moles of ethylene (C2H4) produced. 3. Convert the moles of ethylene to grams using the molar mass of ethylene. 4. Calculate the mass of water (H2O) produced by subtracting the mass of ethylene from the total mass of the ethanol solution.

Let's perform the calculations step by step.

Step 1: Calculate Moles of Oxygen Gas (O2)

To calculate the moles of oxygen gas (O2) required for the combustion reaction, we can use the ideal gas law equation:

PV = nRT

Where: P = pressure (atm) V = volume (L) n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (K)

Given: Volume of oxygen gas (O2) = 25.2 L (n.t.p)

At standard temperature and pressure (n.t.p), the pressure is 1 atm and the temperature is 273 K.

Using the ideal gas law equation, we can calculate the number of moles of oxygen gas (O2):

n = PV / RT

Substituting the values:

n = (1 atm) * (25.2 L) / (0.0821 L·atm/(mol·K) * 273 K)

Calculating the result:

n ≈ 1.12 moles of oxygen gas (O2)

Step 2: Determine Moles of Ethylene (C2H4)

From the balanced chemical equation, we know that 1 mole of ethanol (C2H5OH) reacts to produce 1 mole of ethylene (C2H4).

Therefore, the number of moles of ethylene (C2H4) produced is also 1.12 moles.

Step 3: Convert Moles of Ethylene to Grams

To convert moles of ethylene (C2H4) to grams, we need to use the molar mass of ethylene.

The molar mass of ethylene (C2H4) can be calculated by summing the atomic masses of its constituent elements:

C: 12.01 g/mol H: 1.01 g/mol

Molar mass of ethylene (C2H4) = (2 * 12.01 g/mol) + (4 * 1.01 g/mol) = 28.05 g/mol

Multiplying the number of moles of ethylene by its molar mass:

Mass of ethylene = 1.12 moles * 28.05 g/mol

Calculating the result:

Mass of ethylene ≈ 31.44 grams

Step 4: Calculate Mass of Water

To calculate the mass of water (H2O) produced, we subtract the mass of ethylene from the total mass of the ethanol solution.

Unfortunately, the total mass of the ethanol solution is not provided in the question. Without this information, we cannot accurately calculate the mass of water produced.

Please provide the total mass of the ethanol solution so that we can proceed with the calculation.