Combustion of hydrocarbons such as butane () produces carbon dioxide, a "greenhouse gas." greenhouse gases in the earth's atmosphere can trap the sun's heat, raising the average temperature of the earth. for this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.1. write a balanced chemical equation, including physical state symbols, for the combustion of gaseous butane into gaseous carbon dioxide and gaseous water.2. suppose 0.360 kg of butane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 °c. calculate the volume of carbon dioxide gas that is produced be sure your answer has the correct number of significant digits.

1.

The balanced chemical reaction

2.

The produced volume of carbondioxide is 596.6 L.

Explanation:

1.

The combustion of butane in the presence of atmospheric oxygen to form carbondioxide.

The balanced chemical reaction is as follows.

2.

from the given,

Mass of butane = 0.360 kg = 360 gm

Let's convert the kg into moles

Molar mass of butane = 58.12 g/mol

From the reaction 2 moles of butane gives 8 moles of carbon dioxide.

Hence,

Let's calculate the volume of carbondioxide:

Rearrange the formula is as follows.

n= Number of moles = 24.8 moles

R = Gas constant = 0.0821 atm.L/mol.K

T =  Temperature = 20 c = 20+273 = 293 K

P = Pressure = 1 atm

Substitute the all values in equation (1) then we get  volume of carbondioxide.

Therefore, The produced volume of carbondioxide is 596.6 L.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

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