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.

0.708 L

Explanation:

Start with the balanced chemical equation:

calculate the number of moles of octane:

n(octane)=

From the balanced equation we see that the number of carbon dioxide produced is always 16 times the number of moles of octane available.

Therefore n(carbon dioxide)=

We now treat CO2 as an idal gas, which in reality it behaves like. That allows us to use the following ideal gas equation.

The volume of carbon dioxide gas that is produced from the given amount of octane is

Explanation:

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

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:

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 of volume.

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

By Stoichiometry of the reaction:

of octane gas produces of carbon dioxide gas.

So, 41.216 L of octane gas will produce = of carbon dioxide gas.

Hence, the volume of carbon dioxide gas that is produced from the given amount of octane is