Question

Какой объем кислорода (н.у.) потребуется для полного сжигания 1,15 л этилового спирта (P=0,8г/мл)

Answer 1

C2H5OH+3O2= 2CO2 + 3H2O
масса этилового спирта 1,15*0,8=0,92(г)
объем кислорода 67,2*0,92:46= 1,344 (л)

Answer 2

Calculating Oxygen Volume Required for Complete Combustion of Ethanol

To calculate the volume of oxygen required for the complete combustion of 1.15 L of ethanol (P=0.8 g/mL), we can use the balanced chemical equation for the combustion of ethanol.

The balanced chemical equation for the combustion of ethanol is: C2H5OH + 3O2 → 2CO2 + 3H2O

From the equation, we can see that 1 mole of ethanol reacts with 3 moles of oxygen.

First, we need to convert the volume of ethanol to moles using its density and molar mass.

Step 1: Convert Volume of Ethanol to Mass Using the given density of ethanol (P=0.8 g/mL), we can calculate the mass of 1.15 L of ethanol. Mass = Volume × Density Mass = 1.15 L × 0.8 g/mL

Step 2: Convert Mass of Ethanol to Moles Next, we can convert the mass of ethanol to moles using the molar mass of ethanol. Molar mass of ethanol (C2H5OH) = 46.07 g/mol Moles = Mass / Molar mass

Step 3: Calculate Oxygen Volume Required From the balanced chemical equation, we know that 1 mole of ethanol reacts with 3 moles of oxygen. Therefore, the volume of oxygen required can be calculated based on the moles of ethanol.

Let's perform the calculations:

1. Calculate the mass of 1.15 L of ethanol: Mass = 1.15 L × 0.8 g/mL = 0.92 g

2. Convert the mass of ethanol to moles: Moles = 0.92 g / 46.07 g/mol = 0.02 moles

3. Calculate the volume of oxygen required: 1 mole of ethanol requires 3 moles of oxygen. Therefore, 0.02 moles of ethanol will require 0.02 moles × 3 = 0.06 moles of oxygen.

Using the ideal gas law, we can calculate the volume of oxygen required: PV = nRT V = (nRT) / P Where: n = moles of oxygen R = ideal gas constant (0.0821 L·atm/(K·mol)) T = temperature (in Kelvin) P = pressure (assuming standard pressure, 1 atm)

Let's assume the temperature is 298 K (25°C) and the pressure is 1 atm.

V = (0.06 moles × 0.0821 L·atm/(K·mol) × 298 K) / 1 atm V ≈ 1.48 L

Conclusion: The volume of oxygen required for the complete combustion of 1.15 L of ethanol (P=0.8 g/mL) is approximately 1.48 liters.