Question

Suppose 0.210kg of octane 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. Round your answer to 3 significant digits.

Answer 1

Answer: The volume of carbon dioxide gas that is produced from the given amount of octane is
`3.29* 10^2L`

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of octane = 0.210 kg = 210 g (Conversion factor: 1 kg = 1000 g)

Molar mass of octane = 114.23 g/mol

Putting values in above equation, we get:

`\text{Moles of octane}=(210g)/(114.23g/mol)=1.84mol`

The temperature and pressure conditions given to us are NTP conditions:

At NTP:

1 mole of a gas occupies 22.4 L of volume.

So, 1.84 moles of octane gas will occupy
`22.4* 1.84=41.216L` of volume.

The chemical reaction for the combustion of octane follows the equation:

`2C_8H_(18)+25O_2\rightarrow 16CO_2+18H_2O`

By Stoichiometry of the reaction:

`(2* 22.4L)` of octane gas produces
`(16* 22.4L)` of carbon dioxide gas.

So, 41.216 L of octane gas will produce =
`((16* 22.4)L)/((2* 22.4)L)* 41.216=329.728L=3.29* 10^2L` of carbon dioxide gas.

Hence, the volume of carbon dioxide gas that is produced from the given amount of octane is
`3.29* 10^2L`