Question

A 2.50 L sample of butane gas (C4H10), measured at 22.0
oC and 1.20 atm pressure, is combusted completely and
the carbon dioxide gas collected at the same pressure and
temperature. What volume of CO2 is produced?

Answer 1

Answer:

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Answer 2

The resultant volume of the carbon dioxide gas which is produced by the combustion of butane is 14.4L.

How do we calculate the moles of gas?

Moles of any gas will be calculated by using the ideal gas equation as:

PV = nRT, where according to qustion for butane gas

P = pressure = 1.20 atm

V = volume = 2.50L

n = moles = ?

R = universal gas constant = 0.082 L.atm / K.mol

T = temperature = 22 degree celsius = 295.15 K

On putting values we get

n = (1.20)(2.50) / (0.082)(295.15) = 0.124 moles

Given chemical equation is:

C₄H₁₀ + 13/2O₂ → 4CO₂ + 5H₂O

From the stoichiometry of the reaction, it is clear that,

1 mole of C₄H₁₀ = produces 4 moles of CO₂

0.124 mole of C₄H₁₀ = produces 4×0.124=0.496 moles of CO₂

Now we calculate the volume for the carbon dioxide gas at the given pressure and temperature as:

V = (0.496)(0.082)(295.15) / (1.20) = 14.4 L

Hence required volume of CO₂ is 14.4 L.

To know more about ideal gas equation, visit the below link:

https://brainly.com/question/24236411

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