How much vol. of air will be needed for complete combustion of 10 lt. of ethane?

Question

Answer

35 Lt is the answer. As of how,

2C2H6 + 702 - - - -> 4CO2 + 6H20

so basically 2 moles of ethane requires 7 moles of o2 for complete combustion.

2 mole of ethane = (22.4x2)L space occupied
7 moles of O2 = (22.4×7)L space occupied

44.8L of ethane requires 156.8L of O2
10L of ethane requires?

ans - (10×156.8)÷ 44.8
= 35L

Answer

175 L of air is required

Answer

First of all balance the reaction between ethane and oxygen and then we know at stp 1mole = 22.4 LITRES then by using unitary method we can find the following values

Answer

How much volume of air will be needed for complete combustion of 10 liters of ethane?

To determine the volume of air needed for the complete combustion of ethane, we need to consider the stoichiometry of the reaction. The balanced chemical equation for the combustion of ethane (C2H6) is as follows:

C2H6 + 3.5O2 → 2CO2 + 3H2O

From the equation, we can see that one mole of ethane reacts with 3.5 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.

1. Calculate the number of moles of ethane:
To calculate the number of moles of ethane, we need to use the ideal gas law equation:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Assuming standard temperature and pressure (STP) conditions, which are 1 atm and 273 K respectively, we can use the equation to calculate the number of moles of ethane:

n = PV / RT
= (1 atm) (10 L) / (0.0821 L·atm/mol·K) (273 K)
≈ 0.457 moles

2. Calculate the number of moles of oxygen:
From the balanced equation, we know that 1 mole of ethane reacts with 3.5 moles of oxygen. Therefore, the number of moles of oxygen required for the complete combustion of 0.457 moles of ethane can be calculated as follows:

moles of O2 = 3.5 × moles of C2H6
= 3.5 × 0.457
≈ 1.599 moles

3. Calculate the volume of air:
Air is composed of approximately 21% oxygen by volume. To determine the volume of air needed, we need to find the volume of oxygen required and then divide it by 0.21:

volume of air = volume of O2 / 0.21
= (1.599 moles) (22.4 L/mol) / 0.21
≈ 169.81 L

Therefore, approximately 169.81 liters of air will be needed for the complete combustion of 10 liters of ethane.

In summary:
- The number of moles of ethane is calculated using the ideal gas law equation.
- The number of moles of oxygen required is determined based on the stoichiometry of the reaction.
- The volume of air is calculated by dividing the volume of oxygen by the percentage of oxygen in air.

Answer

35lit..

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