Question

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.

Answer 1

1.

The balanced chemical reaction

`2C_(4)H_(10)(g)+13O_(2)(g)\rightarrow 8CO_(2)(g)+10H_(2)O(g)`

2.

The produced volume of carbondioxide is 596.6 L.

Step-by-step explanation:

1.

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

The balanced chemical reaction is as follows.

`2C_(4)H_(10)(g)+13O_(2)(g)\rightarrow 8CO_(2)(g)+10H_(2)O(g)`

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,

`moles\,of\,CO_(2)=(360* 8)/(2*58.12)=24.8 \,\,moles`

Let's calculate the volume of carbondioxide:

`PV=nRT`

Rearrange the formula is as follows.

`V=(nRT)/(P)...............(1)`

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.

`V=(24.8*0.0821*293)/(1)= 596.6 L`

Therefore, The produced volume of carbondioxide is 596.6 L.