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

Answer:

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.

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,

$moles\,of\,CO_{2}=\frac{360\times 8}{2\times58.12}=24.8 \,\,moles$

Let's calculate the volume of carbondioxide:

$PV=nRT$

Rearrange the formula is as follows.

$V=\frac{nRT}{P}...............(1)$

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.

$V=\frac{24.8\times0.0821\times293}{1}= 596.6 L$

Therefore, The produced volume of carbondioxide is 596.6 L.