Question

Определите объём воздуха, требуемый для сжигания 3 литров пентана

Answer 1

 3 л                          Х л
С5Н12        +         8 О2       =      5СО2      +    6Н2О

n = 1 моль            n = 8 моль

Vm = 22,4 л/моль   Vm = 22,4 л/моль

V = 22,4 л               V = 179,2 л

3 л С5Н12     -    Х л О2

22,4 л С5Н12  -  179,2 л О2

V(O2) = 3 * 179,2 / 22,4 = 24 л

Answer 2

Calculating the Volume of Air Required to Burn 3 Liters of Pentane

To calculate the volume of air required to burn 3 liters of pentane, we can use the balanced chemical equation for the combustion of pentane and the stoichiometry of the reaction.

The balanced chemical equation for the combustion of pentane is:

C5H12 + 8O2 → 5CO2 + 6H2O

This equation shows that 1 mole of pentane (C5H12) reacts with 8 moles of oxygen (O2) to produce 5 moles of carbon dioxide (CO2) and 6 moles of water (H2O).

Now, let's calculate the volume of air required to burn 3 liters of pentane.

First, we need to convert the volume of pentane from liters to moles. Then, we can use the mole ratio from the balanced chemical equation to determine the moles of oxygen required. Finally, we can convert the moles of oxygen to the volume of air at standard conditions.

Calculation Steps:

1. Convert 3 Liters of Pentane to Moles: - The molar volume of a gas at standard conditions (0°C and 1 atm) is 22.4 liters/mol. - Using the molar volume, we can convert the volume of pentane to moles.

   3 liters of pentane * (1 mole / 22.4 liters) = moles of pentane
   

2. Use the Mole Ratio to Determine Moles of Oxygen: - From the balanced chemical equation, we know that 1 mole of pentane reacts with 8 moles of oxygen. - We can use this mole ratio to find the moles of oxygen required.

   Moles of pentane * (8 moles O2 / 1 mole C5H12) = moles of oxygen
   

3. Convert Moles of Oxygen to Volume of Air: - The molar volume of a gas at standard conditions is 22.4 liters/mol. - We can use this molar volume to convert the moles of oxygen to the volume of air.

   Moles of oxygen * (22.4 liters / 1 mole) = volume of air (liters)
   

Result:

The volume of air required to burn 3 liters of pentane can be calculated using the steps above. This will give us the volume of air needed for complete combustion of the given amount of pentane.

Let's proceed with the calculations based on the given information.