calculate the volume of oxygen required for complete combustion of 12.5 dm cube ethane gas at NTP. give me solution of this question

Question

calculate  the volume of oxygen required for complete combustion of 12.5  dm cube ethane gas at NTP.                                give me solution of this question

Answer

Calculation of Volume of Oxygen Required for Complete Combustion of Ethane Gas at NTP

Formula Used:

The balanced chemical equation for the combustion of ethane gas is:

C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O

From the equation, it can be observed that 1 mole of ethane requires 3.5 moles of oxygen for complete combustion.

The volume of oxygen required can be calculated using the ideal gas law:

V = nRT/P

Where,

V = Volume of gas (in dm³)

n = Number of moles of gas

R = Universal gas constant (0.0821 L atm K^-1 mol^-1)

T = Temperature of gas (in Kelvin)

P = Pressure of gas (in atm)

Solution:

Given, volume of ethane gas = 12.5 dm³

At NTP, temperature (T) = 273 K and pressure (P) = 1 atm

First, let's calculate the number of moles of ethane:

1 mole of any gas occupies 22.4 dm³ at NTP

Therefore, number of moles of ethane = volume of ethane gas / 22.4

Number of moles of ethane = 12.5 / 22.4 = 0.558 moles

Now, let's calculate the number of moles of oxygen required for complete combustion of ethane:

Number of moles of oxygen = 3.5 x number of moles of ethane

Number of moles of oxygen = 3.5 x 0.558 = 1.953 moles

Finally, let's calculate the volume of oxygen required:

V = nRT/P

V = (1.953 x 0.0821 x 273) / 1

V = 44.6 dm³

Therefore, the volume of oxygen required for complete combustion of 12.5 dm³ ethane gas at NTP is 44.6 dm³.

Answer

reaction for combustion of Ethane is 2(C2H6) + 7(O2) ---> 4(CO2)+ 6H2Ofor combustion of '''2 ''mole of ethane ''' 7'''  mole oxygen is required. It also means that '''2*(22.4)dmcube''  ethane require ''' 7*(22.4) dmcube''' oxygen2*22.4. --------> 7*22.4 dmcube1.   -----------> (7*22.4)/(2*22.4)=> 1 dmcube {CO2} ------>7/2                                         dmcube {O2}it implies that  '' 12.5 dmcube '' CO2 require.   '12.5*7/2 dmcube ''' O2 which equal to ''' 43.75 dmcube ''' oxygen

Answer

thanks yaar maine tumhara answer aaj dekha

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