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What volume of [tex]C_8H_{18}[/tex] is required to fill a 1.4 dm³ airbag with [tex]CO_2[/tex] if the wrecker truck burns octane at STP?

Answer :

Therefore, the volume of octane required to produce the desired amount of carbon dioxide is 1.23 dm3

The given balanced chemical equation is:

2C8H18 + 25O2 → 16CO2 + 18H2O

The balanced equation states that 2 moles of octane react with 25 moles of oxygen to produce 16 moles of carbon dioxide and 18 moles of water.

The volume of CO2 can be calculated from the volume of octane.

Let's make use of the ideal gas law,

PV = nRT

where P = pressure, V = volume, n = number of moles, R = gas constant and T = temperature.

Volume of C8H18 is given by:

P1V1/T1 = P2V2/T2

The volume of C8H18 required is V1 = P2V2T1/P1T2

The pressure, P = 1 atm, Temperature, T = 273 K, Volume, V2 = 1.4 dm3, R = 0.082 L atm K-1 mol-1

Since 2 moles of octane produce 16 moles of CO2, therefore 1 mole of octane produces 8 moles of CO2.

P1 = pressure of C8H18 = 1 atm, T1 = temperature of C8H18 = 273K

Applying the above values, the volume of octane can be calculated as follows:

V1 = P2V2T1/P1T2

V1 = (1 atm) (1.4 dm3) (273 K)/(1 atm) (298 K)

V1 = 1.23 dm3

to know more about ideal gas law visit:

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