Question

Combustion of hydrocarbons such as octane (C_8H_18) 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 liquid octane into gaseous carbon dioxide and gaseous water.
2. Suppose 0.330 kg of octane are burned in air at a pressure of exactly 1 atm and a temperature of 10.0 ?C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

Answer 1

Answer: 1. $2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)$

2. The volume of carbon dioxide gas that is produced is 537 L

Explanation:

1. Combustion is defined as the type of chemical reaction where a hydrocarbon is combusted in the presence of oxygen to give carbon dioxide and water.

The balanced reaction for combustion of octane is:

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)$

2.

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$   $\text{Moles of octane}=\frac{0.330\times 1000g}{114.23g/mol}=2.89moles$

According to stoichiometry:

2 moles of octane produce = 16 moles of carbon dioxide

2.89 moles of octane produce = $\frac{16}{2}\times 2.89=23.1moles$ of carbon dioxide

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm

V = Volume of gas = ?

n = number of moles = 23.1

R = gas constant =$0.0821Latm/Kmol$

T =temperature =$10.0^0C=(10+273)K=283K$

$V=\frac{nRT}{P}$

$V=\frac{23.1\times 0.0821Latm/K mol\times 283K}{1atm}=537L$

Thus the volume of carbon dioxide gas that is produced is 537 L